OP Malhotra Class 11 Maths Solutions Chapter 17 Circle Exercise 17 (B)

Question 1. Find the centre and radius of the circle
(i) \( x^2 + y^2 + 4x - 4y - 1 = 0 \);
(ii) \( 2x^2 + 2y^2 = 3x - 5y + 7 \).
Answer:
(i) The given equation of the circle is \( x^2 + y^2 + 4x - 4y - 1 = 0 \).
We compare this with the general equation of a circle, which is \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
From the comparison, we find:
\( 2g = 4 \implies g = 2 \)
\( 2f = -4 \implies f = -2 \)
\( c = -1 \)
The centre of the circle is \( (-g, -f) \). So, the centre is \( (-2, -(-2)) = (-2, 2) \).
The radius of the circle is found using the formula \( r = \sqrt{g^2 + f^2 - c} \).
\( r = \sqrt{2^2 + (-2)^2 - (-1)} \)
\( r = \sqrt{4 + 4 + 1} \)
\( r = \sqrt{9} \)
\( r = 3 \)
(ii) The given equation of the circle is \( 2x^2 + 2y^2 = 3x - 5y + 7 \).
First, we need to rewrite this equation into the standard general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \). To do this, we divide the entire equation by 2 and move all terms to one side:
\( x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0 \)
Now, we compare this with the general equation:
\( 2g = -\frac{3}{2} \implies g = -\frac{3}{4} \)
\( 2f = \frac{5}{2} \implies f = \frac{5}{4} \)
\( c = -\frac{7}{2} \)
The centre of the circle is \( (-g, -f) \). So, the centre is \( (-(-\frac{3}{4}), -(\frac{5}{4})) = (\frac{3}{4}, -\frac{5}{4}) \).
The radius of the circle is found using the formula \( r = \sqrt{g^2 + f^2 - c} \).
\( r = \sqrt{(-\frac{3}{4})^2 + (\frac{5}{4})^2 - (-\frac{7}{2})} \)
\( r = \sqrt{\frac{9}{16} + \frac{25}{16} + \frac{7}{2}} \)
To add these fractions, we find a common denominator, which is 16:
\( r = \sqrt{\frac{9}{16} + \frac{25}{16} + \frac{56}{16}} \)
\( r = \sqrt{\frac{9 + 25 + 56}{16}} \)
\( r = \sqrt{\frac{90}{16}} \)
\( r = \frac{\sqrt{90}}{\sqrt{16}} = \frac{\sqrt{9 \times 10}}{4} = \frac{3\sqrt{10}}{4} \)
In simple words: To find the centre and radius of a circle, first make sure its equation looks like \( x^2 + y^2 + 2gx + 2fy + c = 0 \). The centre is then \( (-g, -f) \), and the radius is found using a specific formula. The general form of a circle's equation helps identify these key properties easily.

🎯 Exam Tip: Always remember to transform the given equation into the standard form \( x^2 + y^2 + 2gx + 2fy + c = 0 \) before identifying g, f, and c. Ensure the coefficients of \( x^2 \) and \( y^2 \) are both 1.

Question 2. Find the lengths of the intercepts of the circle \( 3x^2 + 3y^2 - 5x + 3y = 0 \) on the coordinates axes.
Answer:
The given equation of the circle is \( 3x^2 + 3y^2 - 5x + 3y = 0 \).

To find the length of the x-intercept, we set \( y = 0 \) in the circle's equation:
\( 3x^2 + 3(0)^2 - 5x + 3(0) = 0 \)
\( 3x^2 - 5x = 0 \)
Factor out \( x \):
\( x(3x - 5) = 0 \)
This gives two values for \( x \): \( x = 0 \) and \( 3x - 5 = 0 \implies x = \frac{5}{3} \).
The length of the x-intercept is the absolute difference between these two x-values:
Length of x-intercept \( = |\frac{5}{3} - 0| = \frac{5}{3} \).

To find the length of the y-intercept, we set \( x = 0 \) in the circle's equation:
\( 3(0)^2 + 3y^2 - 5(0) + 3y = 0 \)
\( 3y^2 + 3y = 0 \)
Factor out \( 3y \):
\( 3y(y + 1) = 0 \)
This gives two values for \( y \): \( y = 0 \) and \( y + 1 = 0 \implies y = -1 \).
The length of the y-intercept is the absolute difference between these two y-values:
Length of y-intercept \( = |0 - (-1)| = |1| = 1 \).
In simple words: The intercepts are where the circle crosses the x-axis or y-axis. To find the x-intercepts, you put \( y=0 \) into the circle's equation and solve for \( x \). To find the y-intercepts, you put \( x=0 \) and solve for \( y \). The length of an intercept is the distance between these two crossing points.

🎯 Exam Tip: Always remember that x-intercepts are found by setting \( y = 0 \) and y-intercepts by setting \( x = 0 \). The length of an intercept is the distance between the two points where the circle crosses that axis.

Question 3. Find the equation of the circle, which passes through the point (5, 4) and is concentric with the circle \( x^2 + y^2 - 8x - 12y + 15 = 0 \).
Answer:
The equation of the given circle is \( x^2 + y^2 - 8x - 12y + 15 = 0 \).
To find the centre of this circle, we compare it with the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
Here, \( 2g = -8 \implies g = -4 \).
And \( 2f = -12 \implies f = -6 \).
So, the centre of the given circle is \( (-g, -f) = (4, 6) \).

The required circle is concentric with this given circle, which means it shares the same centre. Therefore, the centre of the required circle is also \( (4, 6) \).
The required circle passes through the point \( (5, 4) \).
The radius of the required circle will be the distance between its centre \( (4, 6) \) and the point it passes through \( (5, 4) \).
Using the distance formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \):
Radius \( r = \sqrt{(5-4)^2 + (4-6)^2} \)
\( r = \sqrt{1^2 + (-2)^2} \)
\( r = \sqrt{1 + 4} \)
\( r = \sqrt{5} \)

Now, we can write the equation of the required circle using the standard form \( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h, k) \) is the centre and \( r \) is the radius.
\( (x-4)^2 + (y-6)^2 = (\sqrt{5})^2 \)
\( (x-4)^2 + (y-6)^2 = 5 \)
Expand the squares:
\( x^2 - 8x + 16 + y^2 - 12y + 36 = 5 \)
Combine constant terms and move 5 to the left side:
\( x^2 + y^2 - 8x - 12y + 16 + 36 - 5 = 0 \)
\( x^2 + y^2 - 8x - 12y + 47 = 0 \)
In simple words: If two circles are "concentric", it means they have the exact same middle point (centre). So, we found the centre of the given circle. Then, using this centre and the point the new circle passes through, we calculated how big its radius must be. Finally, we wrote the equation for this new circle. Concentric circles are like rings that share a common center.

🎯 Exam Tip: Remember that "concentric" means sharing the same center. The radius of a circle passing through a point is the distance between its center and that point.

Question 4. The radius of the circle \( x^2 + y^2 - 2x + 3y + k = 0 \) is \( 2\frac{1}{2} \). Find the value of k. Find also the equation of the diameter of the circle, which passes through the point \( (5,2\frac{1}{2}) \).
Answer:
The given equation of the circle is \( x^2 + y^2 - 2x + 3y + k = 0 \).
Comparing this with the general equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
\( 2g = -2 \implies g = -1 \)
\( 2f = 3 \implies f = \frac{3}{2} \)
\( c = k \)
The centre of the circle is \( (-g, -f) = (1, -\frac{3}{2}) \).
The radius of the circle is given as \( 2\frac{1}{2} \), which is \( \frac{5}{2} \).
The formula for the radius is \( r = \sqrt{g^2 + f^2 - c} \).
Substitute the known values:
\( \frac{5}{2} = \sqrt{(-1)^2 + (\frac{3}{2})^2 - k} \)
\( \frac{5}{2} = \sqrt{1 + \frac{9}{4} - k} \)
To solve for \( k \), square both sides of the equation:
\( (\frac{5}{2})^2 = 1 + \frac{9}{4} - k \)
\( \frac{25}{4} = \frac{4}{4} + \frac{9}{4} - k \)
\( \frac{25}{4} = \frac{13}{4} - k \)
Now, we can find \( k \):
\( k = \frac{13}{4} - \frac{25}{4} \)
\( k = \frac{13 - 25}{4} \)
\( k = \frac{-12}{4} \)
\( k = -3 \)

Next, we need to find the equation of the diameter that passes through the point \( (5, 2\frac{1}{2}) \).
The point is \( (5, \frac{5}{2}) \).
A diameter always passes through the centre of the circle, which is \( (1, -\frac{3}{2}) \).
So, we need the equation of a line passing through \( (1, -\frac{3}{2}) \) and \( (5, \frac{5}{2}) \).
First, calculate the slope \( m \) of this line:
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{5}{2} - (-\frac{3}{2})}{5 - 1} \)
\( m = \frac{\frac{5+3}{2}}{4} = \frac{\frac{8}{2}}{4} = \frac{4}{4} = 1 \)
Now, use the point-slope form of a linear equation \( y - y_1 = m(x - x_1) \). We can use the centre \( (1, -\frac{3}{2}) \):
\( y - (-\frac{3}{2}) = 1(x - 1) \)
\( y + \frac{3}{2} = x - 1 \)
To remove the fraction, multiply the entire equation by 2:
\( 2(y + \frac{3}{2}) = 2(x - 1) \)
\( 2y + 3 = 2x - 2 \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( 2x - 2y - 2 - 3 = 0 \)
\( 2x - 2y - 5 = 0 \)
In simple words: We used the given radius and the circle's equation to find the missing value 'k'. Then, we found the circle's centre. Because a diameter always goes through the centre, we used the centre point and the other given point to draw the line that represents the diameter. The diameter is a straight line that cuts the circle exactly in half.

🎯 Exam Tip: Remember the formula for the radius \( r = \sqrt{g^2 + f^2 - c} \) and the centre \( (-g, -f) \). Also, know how to find the equation of a line passing through two points using the slope-intercept or point-slope form.

Question 5. Prove that the circle \( x^2 + y^2 - 6x - 2y + 9 = 0 \) (i) touches the x-axis ; (ii) lies entirely inside the circle \( x^2 + y^2 = 18 \).
Answer:
(i) The equation of the given circle is \( x^2 + y^2 - 6x - 2y + 9 = 0 \).
To check if the circle touches the x-axis, we set \( y = 0 \) in its equation:
\( x^2 + (0)^2 - 6x - 2(0) + 9 = 0 \)
\( x^2 - 6x + 9 = 0 \)
This is a perfect square trinomial, which can be factored as:
\( (x - 3)^2 = 0 \)
This equation gives a repeated root: \( x = 3 \).
Since there is only one distinct x-value where the circle intersects the x-axis, the circle touches the x-axis at the point \( (3, 0) \).
In simple words: When a circle just "touches" an axis, it means it meets that axis at only one single point. We check this by putting \( y=0 \) for the x-axis (or \( x=0 \) for the y-axis) into the circle's equation. If the resulting quadratic equation has only one answer (a repeated root), then it touches the axis.

(ii) For the first circle, \( x^2 + y^2 - 6x - 2y + 9 = 0 \):
Comparing with \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
\( 2g_1 = -6 \implies g_1 = -3 \)
\( 2f_1 = -2 \implies f_1 = -1 \)
\( c_1 = 9 \)
The centre \( C_1 = (-g_1, -f_1) = (3, 1) \).
The radius \( r_1 = \sqrt{g_1^2 + f_1^2 - c_1} = \sqrt{(-3)^2 + (-1)^2 - 9} = \sqrt{9 + 1 - 9} = \sqrt{1} = 1 \).

For the second circle, \( x^2 + y^2 = 18 \):
This equation is in the form \( x^2 + y^2 = R^2 \).
The centre \( C_2 = (0, 0) \).
The radius \( r_2 = \sqrt{18} = 3\sqrt{2} \approx 3 \times 1.414 = 4.242 \).

Now, we find the distance between the two centres \( C_1(3, 1) \) and \( C_2(0, 0) \).
Distance \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{(0-3)^2 + (0-1)^2} \)
\( d = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \approx 3.162 \).

For one circle to lie entirely inside another, the distance between their centres plus the radius of the inner circle must be less than the radius of the outer circle (i.e., \( d + r_1 < r_2 \)).
Let's check this condition:
\( \sqrt{10} + 1 < \sqrt{18} \)
\( 3.162 + 1 < 4.242 \)
\( 4.162 < 4.242 \)
Since this condition is true, the circle \( x^2 + y^2 - 6x - 2y + 9 = 0 \) lies entirely inside the circle \( x^2 + y^2 = 18 \).
In simple words: We found the centre and size (radius) of both circles. Then, we measured the distance between their middle points. If the distance between the centres, plus the radius of the smaller circle, is less than the radius of the bigger circle, it means the smaller one fits completely inside the bigger one. When a circle is completely inside another, the distance between their centers plus the inner radius must be less than the outer radius.

🎯 Exam Tip: To prove a circle touches an axis, set the opposite coordinate to zero and show a repeated root. To show one circle lies entirely within another, compare the distance between centers \( d \) with the radii \( r_1 \) and \( r_2 \): the condition is \( d + r_{\text{smaller}} < r_{\text{larger}} \).

Question 6. Find the co-ordinates of the centre of the circle \( x^2 + y^2 - 4x + 6y = 3 \). Given that the point A, outside the circle, has coordinates (a, b) where a and b are both positive, and that the tangents drawn from A to the circle are parallel to the two axes respectively, find the values of a and b.
Answer:
The given equation of the circle is \( x^2 + y^2 - 4x + 6y = 3 \).
First, we rewrite it in the standard general form by moving the constant term to the left side:
\( x^2 + y^2 - 4x + 6y - 3 = 0 \)
Comparing this with \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
\( 2g = -4 \implies g = -2 \)
\( 2f = 6 \implies f = 3 \)
\( c = -3 \)
The centre of the circle \( C = (-g, -f) = (2, -3) \).
The radius of the circle \( r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + 3^2 - (-3)} \)
\( r = \sqrt{4 + 9 + 3} = \sqrt{16} = 4 \).

Now, we are given that point A has coordinates \( (a, b) \) where \( a \) and \( b \) are both positive. The tangents drawn from A to the circle are parallel to the two coordinate axes.
This means that the lines \( x = a \) and \( y = b \) are tangents to the circle.
For the line \( x = a \) to be a tangent, the perpendicular distance from the centre \( (2, -3) \) to this line must be equal to the radius \( 4 \).
The distance from \( (h, k) \) to \( x=a \) is \( |h-a| \). So, \( |2 - a| = 4 \).
This gives two possibilities:
\( 2 - a = 4 \implies a = 2 - 4 = -2 \)
OR
\( 2 - a = -4 \implies a = 2 + 4 = 6 \)

For the line \( y = b \) to be a tangent, the perpendicular distance from the centre \( (2, -3) \) to this line must be equal to the radius \( 4 \).
The distance from \( (h, k) \) to \( y=b \) is \( |k-b| \). So, \( |-3 - b| = 4 \).
This gives two possibilities:
\( -3 - b = 4 \implies b = -3 - 4 = -7 \)
OR
\( -3 - b = -4 \implies b = -3 + 4 = 1 \)

We are given that \( a \) and \( b \) must both be positive.
From the possible values for \( a \), only \( a = 6 \) is positive.
From the possible values for \( b \), only \( b = 1 \) is positive.
Therefore, the values are \( a = 6 \) and \( b = 1 \).
In simple words: First, we found the middle point and size of the circle. We know that if a line just touches a circle (is a tangent), the distance from the circle's middle point to that line is exactly its radius. Since the tangents were straight up-and-down or side-to-side lines (parallel to axes), we used this distance rule to find the 'a' and 'b' values for point A, making sure they were positive. The distance from the center of a circle to any tangent line is always equal to its radius.

🎯 Exam Tip: When tangents are parallel to the coordinate axes, their equations are of the form \( x = \text{constant} \) or \( y = \text{constant} \). The distance from the circle's centre \( (h,k) \) to these lines \( x=a \) or \( y=b \) is simply \( |h-a| \) or \( |k-b| \) respectively, which must equal the radius.

Question 7. Find the equation of the circle whose centre is at the point (4, 5) and which touches the x-axis. Also find the coordinates of the points at which the circle cuts the y-axis.
Answer:
The centre of the circle is given as \( C = (4, 5) \).
Since the circle touches the x-axis, its radius is the absolute value of the y-coordinate of the centre.
So, radius \( r = |5| = 5 \).
The equation of a circle with centre \( (h, k) \) and radius \( r \) is \( (x-h)^2 + (y-k)^2 = r^2 \).
Substitute the centre \( (4, 5) \) and radius \( 5 \):
\( (x-4)^2 + (y-5)^2 = 5^2 \)
\( (x-4)^2 + (y-5)^2 = 25 \)
Expand the terms to get the general form:
\( x^2 - 8x + 16 + y^2 - 10y + 25 = 25 \)
Move all terms to one side:
\( x^2 + y^2 - 8x - 10y + 16 + 25 - 25 = 0 \)
\( x^2 + y^2 - 8x - 10y + 16 = 0 \). This is the equation of the circle.

Next, we need to find the coordinates of the points where the circle cuts the y-axis.
To find the y-intercepts, we set \( x = 0 \) in the circle's equation:
\( (0)^2 + y^2 - 8(0) - 10y + 16 = 0 \)
\( y^2 - 10y + 16 = 0 \)
We can factor this quadratic equation:
\( (y - 2)(y - 8) = 0 \)
This gives two possible values for \( y \):
\( y - 2 = 0 \implies y = 2 \)
\( y - 8 = 0 \implies y = 8 \)
So, the circle cuts the y-axis at the points \( (0, 2) \) and \( (0, 8) \).
In simple words: We know the circle's middle point. Since it just touches the x-axis, its radius is simply the height of its middle point. Once we had the centre and radius, we wrote down the circle's equation. Then, to find where it crosses the y-axis, we imagined the x-value to be zero and solved for the y-values. When a circle touches the x-axis, its radius is simply the absolute value of the y-coordinate of its center.

🎯 Exam Tip: If a circle touches the x-axis, its radius is \( |k| \) (absolute value of y-coordinate of centre). If it touches the y-axis, its radius is \( |h| \) (absolute value of x-coordinate of centre).

Question 8. Show that the circles \( x^2 + y^2 - 4x + 6y + 8 = 0 \) and \( x^2 + y^2 - 10x - 6y + 14 = 0 \) touch at the point \( (3, -1) \).
Answer:
For the first circle, let its equation be \( C_1: x^2 + y^2 - 4x + 6y + 8 = 0 \).
Comparing with the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
\( 2g_1 = -4 \implies g_1 = -2 \)
\( 2f_1 = 6 \implies f_1 = 3 \)
\( c_1 = 8 \)
The centre of the first circle is \( C_1 = (-g_1, -f_1) = (2, -3) \).
The radius of the first circle is \( r_1 = \sqrt{g_1^2 + f_1^2 - c_1} = \sqrt{(-2)^2 + 3^2 - 8} = \sqrt{4 + 9 - 8} = \sqrt{5} \).

For the second circle, let its equation be \( C_2: x^2 + y^2 - 10x - 6y + 14 = 0 \).
Comparing with the general form:
\( 2g_2 = -10 \implies g_2 = -5 \)
\( 2f_2 = -6 \implies f_2 = -3 \)
\( c_2 = 14 \)
The centre of the second circle is \( C_2 = (-g_2, -f_2) = (5, 3) \).
The radius of the second circle is \( r_2 = \sqrt{g_2^2 + f_2^2 - c_2} = \sqrt{(-5)^2 + (-3)^2 - 14} = \sqrt{25 + 9 - 14} = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \).

Now, calculate the distance between the centres \( C_1(2, -3) \) and \( C_2(5, 3) \).
Distance \( d = C_1C_2 = \sqrt{(5-2)^2 + (3-(-3))^2} = \sqrt{3^2 + (3+3)^2} \)
\( d = \sqrt{9 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \).

Next, check the sum of the radii:
\( r_1 + r_2 = \sqrt{5} + 2\sqrt{5} = 3\sqrt{5} \).
Since the distance between the centres \( C_1C_2 \) is equal to the sum of their radii \( r_1 + r_2 \), the two circles touch each other externally.

To show that they touch at the point \( (3, -1) \), we can use the section formula. The point of contact P divides the line segment \( C_1C_2 \) in the ratio \( r_1:r_2 \) internally.
Ratio \( m_1:m_2 = r_1:r_2 = \sqrt{5} : 2\sqrt{5} = 1:2 \).
Using the section formula \( P(x, y) = \left(\frac{m_1x_2 + m_2x_1}{m_1+m_2}, \frac{m_1y_2 + m_2y_1}{m_1+m_2}\right) \):
\( x = \frac{1(5) + 2(2)}{1+2} = \frac{5+4}{3} = \frac{9}{3} = 3 \)
\( y = \frac{1(3) + 2(-3)}{1+2} = \frac{3-6}{3} = \frac{-3}{3} = -1 \)
So, the point of contact is \( (3, -1) \), which matches the given point.
In simple words: We found the middle points and sizes (radii) of both circles. We then measured the distance between their centres. Since this distance was exactly equal to the sum of their radii, it means the circles just touch each other from the outside. Then, using a special formula, we found the exact spot where they touch, and it was the point given in the question. The point where two circles touch externally lies on the line connecting their centers.

🎯 Exam Tip: Circles touch externally if the distance between their centers equals the sum of their radii \( (C_1C_2 = r_1 + r_2) \). The point of contact can be found using the section formula, dividing the line segment joining the centers in the ratio of their radii.

Question 9. Show that the circles \( x^2 + y^2 - 2x = 0 \) and \( x^2 + y^2 + 6x - 6y + 2 = 0 \) touch externally at the point \( (\frac{1}{5}, \frac{3}{5}) \).
Answer:
For the first circle, let its equation be \( C_1: x^2 + y^2 - 2x = 0 \).
Comparing with the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
\( 2g_1 = -2 \implies g_1 = -1 \)
\( 2f_1 = 0 \implies f_1 = 0 \)
\( c_1 = 0 \)
The centre of the first circle is \( C_1 = (-g_1, -f_1) = (1, 0) \).
The radius of the first circle is \( r_1 = \sqrt{g_1^2 + f_1^2 - c_1} = \sqrt{(-1)^2 + 0^2 - 0} = \sqrt{1} = 1 \).

For the second circle, let its equation be \( C_2: x^2 + y^2 + 6x - 6y + 2 = 0 \).
Comparing with the general form:
\( 2g_2 = 6 \implies g_2 = 3 \)
\( 2f_2 = -6 \implies f_2 = -3 \)
\( c_2 = 2 \)
The centre of the second circle is \( C_2 = (-g_2, -f_2) = (-3, 3) \).
The radius of the second circle is \( r_2 = \sqrt{g_2^2 + f_2^2 - c_2} = \sqrt{3^2 + (-3)^2 - 2} = \sqrt{9 + 9 - 2} = \sqrt{16} = 4 \).

Now, calculate the distance between the centres \( C_1(1, 0) \) and \( C_2(-3, 3) \).
Distance \( d = C_1C_2 = \sqrt{(-3-1)^2 + (3-0)^2} = \sqrt{(-4)^2 + 3^2} \)
\( d = \sqrt{16 + 9} = \sqrt{25} = 5 \).

Next, check the sum of the radii:
\( r_1 + r_2 = 1 + 4 = 5 \).
Since the distance between the centres \( C_1C_2 \) is equal to the sum of their radii \( r_1 + r_2 \), the two circles touch each other externally.

To show that they touch at the point \( (\frac{1}{5}, \frac{3}{5}) \), we use the section formula. The point of contact P divides the line segment \( C_1C_2 \) in the ratio \( r_1:r_2 \) internally.
Ratio \( m_1:m_2 = r_1:r_2 = 1:4 \).
Using the section formula \( P(x, y) = \left(\frac{m_1x_2 + m_2x_1}{m_1+m_2}, \frac{m_1y_2 + m_2y_1}{m_1+m_2}\right) \):
\( x = \frac{1(-3) + 4(1)}{1+4} = \frac{-3+4}{5} = \frac{1}{5} \)
\( y = \frac{1(3) + 4(0)}{1+4} = \frac{3+0}{5} = \frac{3}{5} \)
So, the point of contact is \( (\frac{1}{5}, \frac{3}{5}) \), which matches the given point.
In simple words: We calculated the middle points and sizes of both circles. The distance between their middle points was exactly the same as adding their radii together, which means they touch from the outside. Then, using a formula for points that divide a line, we confirmed that they touch at the exact spot mentioned in the question. External touching means the circles meet at one point without overlapping, with the distance between centers equalling the sum of their radii.

🎯 Exam Tip: When two circles touch externally, the distance between their centers is equal to the sum of their radii. The point of contact will divide the line segment joining their centers internally in the ratio of their radii.

Question 10. Show that the circles \( x^2 + y^2 + 2x - 6y + 9 = 0 \) and \( x^2 + y^2 + 8x - 6y + 9 = 0 \) touch internally.
Answer:
For the first circle, let its equation be \( C_1: x^2 + y^2 + 2x - 6y + 9 = 0 \).
Comparing with the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
\( 2g_1 = 2 \implies g_1 = 1 \)
\( 2f_1 = -6 \implies f_1 = -3 \)
\( c_1 = 9 \)
The centre of the first circle is \( C_1 = (-g_1, -f_1) = (-1, 3) \).
The radius of the first circle is \( r_1 = \sqrt{g_1^2 + f_1^2 - c_1} = \sqrt{1^2 + (-3)^2 - 9} = \sqrt{1 + 9 - 9} = \sqrt{1} = 1 \).

For the second circle, let its equation be \( C_2: x^2 + y^2 + 8x - 6y + 9 = 0 \).
Comparing with the general form:
\( 2g_2 = 8 \implies g_2 = 4 \)
\( 2f_2 = -6 \implies f_2 = -3 \)
\( c_2 = 9 \)
The centre of the second circle is \( C_2 = (-g_2, -f_2) = (-4, 3) \).
The radius of the second circle is \( r_2 = \sqrt{g_2^2 + f_2^2 - c_2} = \sqrt{4^2 + (-3)^2 - 9} = \sqrt{16 + 9 - 9} = \sqrt{16} = 4 \).

Now, calculate the distance between the centres \( C_1(-1, 3) \) and \( C_2(-4, 3) \).
Distance \( d = C_1C_2 = \sqrt{(-4-(-1))^2 + (3-3)^2} = \sqrt{(-4+1)^2 + 0^2} \)
\( d = \sqrt{(-3)^2} = \sqrt{9} = 3 \).

Next, check the absolute difference of the radii:
\( |r_1 - r_2| = |1 - 4| = |-3| = 3 \).
Since the distance between the centres \( C_1C_2 \) is equal to the absolute difference of their radii \( |r_1 - r_2| \), the two circles touch each other internally.
In simple words: We found the middle points and sizes (radii) of both circles. Then, we measured the distance between their centres. Since this distance was exactly the same as the difference between their radii, it means one circle is inside the other and they just touch at one point. When circles touch internally, one circle is inside the other, and the distance between their centers equals the difference of their radii.

🎯 Exam Tip: Circles touch internally if the distance between their centers equals the absolute difference of their radii \( (C_1C_2 = |r_1 - r_2|) \).

Question 11. Find the equation of the circle which passes through the points (0, 0), (0, 1) and (2, 3).
Answer:
Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

The circle passes through the point \( (0, 0) \). Substitute \( x=0, y=0 \) into the equation:
\( (0)^2 + (0)^2 + 2g(0) + 2f(0) + c = 0 \)
\( 0 + 0 + 0 + 0 + c = 0 \implies c = 0 \).

The circle passes through the point \( (0, 1) \). Substitute \( x=0, y=1 \) and \( c=0 \) into the equation:
\( (0)^2 + (1)^2 + 2g(0) + 2f(1) + 0 = 0 \)
\( 1 + 0 + 2f = 0 \implies 2f = -1 \implies f = -\frac{1}{2} \).

The circle passes through the point \( (2, 3) \). Substitute \( x=2, y=3 \) and the values of \( f \) and \( c \) into the equation:
\( (2)^2 + (3)^2 + 2g(2) + 2f(3) + c = 0 \)
\( 4 + 9 + 4g + 6f + c = 0 \)
\( 13 + 4g + 6(-\frac{1}{2}) + 0 = 0 \)
\( 13 + 4g - 3 = 0 \)
\( 10 + 4g = 0 \implies 4g = -10 \implies g = -\frac{10}{4} = -\frac{5}{2} \).

Now we have the values for \( g, f, \) and \( c \):
\( g = -\frac{5}{2} \)
\( f = -\frac{1}{2} \)
\( c = 0 \)
Substitute these values back into the general equation of the circle:
\( x^2 + y^2 + 2(-\frac{5}{2})x + 2(-\frac{1}{2})y + 0 = 0 \)
\( x^2 + y^2 - 5x - y = 0 \).
This is the required equation of the circle.
In simple words: We used the general formula for a circle's equation. Since we know three points the circle passes through, we put each point's coordinates into the formula. This gave us three small equations. We solved these equations step-by-step to find the hidden numbers (g, f, c) that define the circle, and then we put them back into the main formula. Three non-collinear points uniquely determine a circle, meaning there's only one circle that can pass through all three.

🎯 Exam Tip: When given three points, always start with the general equation of a circle. Substituting the points will give you a system of linear equations in g, f, and c, which you then solve to find the specific equation.

Question 12. Find the centre and radius of the circle which passes through the points (7, 5), (6, -2), (-1, -1).
Answer:
Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
Since the circle passes through the points (7, 5), (6, -2), and (-1, -1), we substitute each point's coordinates into the general equation:

For point \( (7, 5) \):
\( 7^2 + 5^2 + 2g(7) + 2f(5) + c = 0 \)
\( 49 + 25 + 14g + 10f + c = 0 \)
\( 14g + 10f + c + 74 = 0 \) (Equation 2)

For point \( (6, -2) \):
\( 6^2 + (-2)^2 + 2g(6) + 2f(-2) + c = 0 \)
\( 36 + 4 + 12g - 4f + c = 0 \)
\( 12g - 4f + c + 40 = 0 \) (Equation 3)

For point \( (-1, -1) \):
\( (-1)^2 + (-1)^2 + 2g(-1) + 2f(-1) + c = 0 \)
\( 1 + 1 - 2g - 2f + c = 0 \)
\( -2g - 2f + c + 2 = 0 \) (Equation 4)

Now, we solve these three linear equations for \( g, f, \) and \( c \):
Subtract Equation 3 from Equation 2:
\( (14g + 10f + c + 74) - (12g - 4f + c + 40) = 0 \)
\( 2g + 14f + 34 = 0 \)
Divide by 2: \( g + 7f + 17 = 0 \) (Equation 5)

Subtract Equation 4 from Equation 3:
\( (12g - 4f + c + 40) - (-2g - 2f + c + 2) = 0 \)
\( 14g - 2f + 38 = 0 \)
Divide by 2: \( 7g - f + 19 = 0 \) (Equation 6)

From Equation 6, we can express \( f \) in terms of \( g \):
\( f = 7g + 19 \)
Substitute this expression for \( f \) into Equation 5:
\( g + 7(7g + 19) + 17 = 0 \)
\( g + 49g + 133 + 17 = 0 \)
\( 50g + 150 = 0 \)
\( 50g = -150 \implies g = -3 \).

Now, find \( f \) using \( f = 7g + 19 \):
\( f = 7(-3) + 19 = -21 + 19 = -2 \).

Finally, find \( c \) using Equation 4:
\( -2g - 2f + c + 2 = 0 \)
\( -2(-3) - 2(-2) + c + 2 = 0 \)
\( 6 + 4 + c + 2 = 0 \)
\( 12 + c = 0 \implies c = -12 \).

So, we have \( g = -3, f = -2, \) and \( c = -12 \).
The equation of the circle is \( x^2 + y^2 + 2(-3)x + 2(-2)y + (-12) = 0 \), which simplifies to:
\( x^2 + y^2 - 6x - 4y - 12 = 0 \).

The centre of the circle is \( (-g, -f) = (-(-3), -(-2)) = (3, 2) \).
The radius of the circle is \( r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + (-2)^2 - (-12)} \)
\( r = \sqrt{9 + 4 + 12} = \sqrt{25} = 5 \).
In simple words: We used the general equation of a circle and plugged in each of the three given points. This created a set of three equations with three unknowns (g, f, c). We solved these equations step-by-step to find the values for g, f, and c. Once we had these values, we could write the full equation of the circle and then easily find its centre and radius using the standard formulas. Solving a system of three linear equations is a common method to find the specific values (g, f, c) that define the circle's equation.

🎯 Exam Tip: When given three points, you will always set up and solve a system of three linear equations for g, f, and c. Be meticulous with your algebraic manipulation to avoid errors.

Question 13. Find the equation of the circle circumscribing the triangle formed by the lines \( x + y + 1 = 0 \), \( 3x + y - 5 = 0 \) and \( 2x + y - 4 = 0 \).
Answer:
A circle circumscribing a triangle passes through all three vertices of the triangle. So, first, we need to find the coordinates of the vertices of the triangle formed by the given lines:
Line 1: \( x + y + 1 = 0 \)
Line 2: \( 3x + y - 5 = 0 \)
Line 3: \( 2x + y - 4 = 0 \)

**1. Find Vertex A (intersection of Line 1 and Line 2):**
Subtract Line 1 from Line 2:
\( (3x + y - 5) - (x + y + 1) = 0 \)
\( 2x - 6 = 0 \implies 2x = 6 \implies x = 3 \)
Substitute \( x=3 \) into Line 1: \( 3 + y + 1 = 0 \implies y + 4 = 0 \implies y = -4 \).
So, Vertex A is \( (3, -4) \).

**2. Find Vertex B (intersection of Line 2 and Line 3):**
Subtract Line 3 from Line 2:
\( (3x + y - 5) - (2x + y - 4) = 0 \)
\( x - 1 = 0 \implies x = 1 \)
Substitute \( x=1 \) into Line 3: \( 2(1) + y - 4 = 0 \implies 2 + y - 4 = 0 \implies y - 2 = 0 \implies y = 2 \).
So, Vertex B is \( (1, 2) \).

**3. Find Vertex C (intersection of Line 1 and Line 3):**
Subtract Line 1 from Line 3:
\( (2x + y - 4) - (x + y + 1) = 0 \)
\( x - 5 = 0 \implies x = 5 \)
Substitute \( x=5 \) into Line 1: \( 5 + y + 1 = 0 \implies y + 6 = 0 \implies y = -6 \).
So, Vertex C is \( (5, -6) \).

Now, we need to find the equation of the circle that passes through the three points A(3, -4), B(1, 2), and C(5, -6). (This is the same method as Question 12).
Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

For point \( (3, -4) \):
\( 3^2 + (-4)^2 + 2g(3) + 2f(-4) + c = 0 \)
\( 9 + 16 + 6g - 8f + c = 0 \implies 6g - 8f + c + 25 = 0 \) (Equation 5)

For point \( (1, 2) \):
\( 1^2 + 2^2 + 2g(1) + 2f(2) + c = 0 \)
\( 1 + 4 + 2g + 4f + c = 0 \implies 2g + 4f + c + 5 = 0 \) (Equation 6)

For point \( (5, -6) \):
\( 5^2 + (-6)^2 + 2g(5) + 2f(-6) + c = 0 \)
\( 25 + 36 + 10g - 12f + c = 0 \implies 10g - 12f + c + 61 = 0 \) (Equation 7)

Subtract Equation 6 from Equation 5:
\( (6g - 8f + c + 25) - (2g + 4f + c + 5) = 0 \)
\( 4g - 12f + 20 = 0 \)
Divide by 4: \( g - 3f + 5 = 0 \) (Equation 8)

Subtract Equation 6 from Equation 7:
\( (10g - 12f + c + 61) - (2g + 4f + c + 5) = 0 \)
\( 8g - 16f + 56 = 0 \)
Divide by 8: \( g - 2f + 7 = 0 \) (Equation 9)

Subtract Equation 9 from Equation 8:
\( (g - 3f + 5) - (g - 2f + 7) = 0 \)
\( -f - 2 = 0 \implies f = -2 \).

Substitute \( f=-2 \) into Equation 9:
\( g - 2(-2) + 7 = 0 \)
\( g + 4 + 7 = 0 \implies g + 11 = 0 \implies g = -11 \).

Substitute \( g=-11 \) and \( f=-2 \) into Equation 6:
\( 2(-11) + 4(-2) + c + 5 = 0 \)
\( -22 - 8 + c + 5 = 0 \)
\( -25 + c = 0 \implies c = 25 \).

So, we have \( g = -11, f = -2, \) and \( c = 25 \).
Substitute these values into the general equation of the circle:
\( x^2 + y^2 + 2(-11)x + 2(-2)y + 25 = 0 \)
\( x^2 + y^2 - 22x - 4y + 25 = 0 \).
This is the required equation of the circle circumscribing the triangle.
In simple words: First, we found the three corner points (vertices) of the triangle by solving the equations of the lines in pairs. Once we had these three points, we used the same method as in the previous problems: we plugged each point into the general circle equation to get three equations, which we then solved to find the circle's specific formula. A circumcircle is unique for any given triangle, meaning it's the only circle that touches all three corners.

🎯 Exam Tip: The key to solving problems about circumcircles of triangles formed by lines is to first find the vertices of the triangle by solving the line equations simultaneously, then treat it as a problem of finding a circle through three given points.

Question 14. Show that the circle \( x^2 + y^2 - 4x + 4y + 4 = 0 \) touches the co-ordinate axes. If the points of contact are A and B, find the equation of the circle which passes through A, B and the origin.
Answer:
The given equation of the circle is \( x^2 + y^2 - 4x + 4y + 4 = 0 \).

**1. Show that the circle touches the x-axis:**
Set \( y = 0 \) in the equation:
\( x^2 + (0)^2 - 4x + 4(0) + 4 = 0 \)
\( x^2 - 4x + 4 = 0 \)
This is a perfect square trinomial:
\( (x - 2)^2 = 0 \)
This gives a repeated root \( x = 2 \). Since there is only one x-intercept, the circle touches the x-axis at point A \( (2, 0) \).

**2. Show that the circle touches the y-axis:**
Set \( x = 0 \) in the equation:
\( (0)^2 + y^2 - 4(0) + 4y + 4 = 0 \)
\( y^2 + 4y + 4 = 0 \)
This is also a perfect square trinomial:
\( (y + 2)^2 = 0 \)
This gives a repeated root \( y = -2 \). Since there is only one y-intercept, the circle touches the y-axis at point B \( (0, -2) \).

Since the circle touches both the x-axis at \( (2, 0) \) and the y-axis at \( (0, -2) \), it touches the coordinate axes.

**3. Find the equation of the circle which passes through A, B and the origin:**
We need to find the equation of a new circle passing through three points:
Origin \( O = (0, 0) \)
Point A \( = (2, 0) \)
Point B \( = (0, -2) \)
Let the general equation of this new circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

For point \( (0, 0) \):
\( (0)^2 + (0)^2 + 2g(0) + 2f(0) + c = 0 \implies c = 0 \).

For point \( (2, 0) \):
\( (2)^2 + (0)^2 + 2g(2) + 2f(0) + c = 0 \)
\( 4 + 4g + 0 = 0 \implies 4g = -4 \implies g = -1 \).

For point \( (0, -2) \):
\( (0)^2 + (-2)^2 + 2g(0) + 2f(-2) + c = 0 \)
\( 4 - 4f + 0 = 0 \implies 4f = 4 \implies f = 1 \).

Substitute the values \( g = -1, f = 1, \) and \( c = 0 \) into the general equation:
\( x^2 + y^2 + 2(-1)x + 2(1)y + 0 = 0 \)
\( x^2 + y^2 - 2x + 2y = 0 \).
This is the equation of the circle passing through A, B, and the origin.
In simple words: First, we showed that the given circle touches the x-axis (at A) and the y-axis (at B) by proving that when we set one coordinate to zero, the other coordinate's equation has only one answer. Then, we used these two contact points (A and B) plus the origin (0,0) as three points that define a new circle. We then found the equation of this new circle. When a circle touches both coordinate axes, its radius will be equal to the absolute value of both its center's x and y coordinates (in the first quadrant).

🎯 Exam Tip: A repeated root in the quadratic equation (after setting \( x=0 \) or \( y=0 \)) signifies tangency to an axis. The origin \( (0,0) \) simplifies finding the constant term \( c \) in the general equation of a circle.

Question 15. Find the equation of the circle which passes through the points P(1, 0), Q(3, 0) and R(0, 2). Find also (i) the co-ordinates of the other point in which the axis of y cuts the circle, (ii) the coordinates of the other end of the diameter through Q.
Answer:
**Part 1: Find the equation of the circle**
Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

For point P(1, 0):
\( 1^2 + 0^2 + 2g(1) + 2f(0) + c = 0 \implies 1 + 2g + c = 0 \) (Equation 2)

For point Q(3, 0):
\( 3^2 + 0^2 + 2g(3) + 2f(0) + c = 0 \implies 9 + 6g + c = 0 \) (Equation 3)

For point R(0, 2):
\( 0^2 + 2^2 + 2g(0) + 2f(2) + c = 0 \implies 4 + 4f + c = 0 \) (Equation 4)

Subtract Equation 2 from Equation 3:
\( (9 + 6g + c) - (1 + 2g + c) = 0 \)
\( 8 + 4g = 0 \implies 4g = -8 \implies g = -2 \).

Substitute \( g=-2 \) into Equation 2:
\( 1 + 2(-2) + c = 0 \implies 1 - 4 + c = 0 \implies -3 + c = 0 \implies c = 3 \).

Substitute \( c=3 \) into Equation 4:
\( 4 + 4f + 3 = 0 \implies 7 + 4f = 0 \implies 4f = -7 \implies f = -\frac{7}{4} \).

So, we have \( g = -2, f = -\frac{7}{4}, \) and \( c = 3 \).
Substitute these into the general equation:
\( x^2 + y^2 + 2(-2)x + 2(-\frac{7}{4})y + 3 = 0 \)
\( x^2 + y^2 - 4x - \frac{7}{2}y + 3 = 0 \)
To remove the fraction, multiply the entire equation by 2:
\( 2x^2 + 2y^2 - 8x - 7y + 6 = 0 \). This is the equation of the circle.

**Part (i): Co-ordinates of the other point in which the y-axis cuts the circle**
To find where the circle cuts the y-axis, set \( x = 0 \) in the circle's equation:
\( 2(0)^2 + 2y^2 - 8(0) - 7y + 6 = 0 \)
\( 2y^2 - 7y + 6 = 0 \)
Factor the quadratic equation:
\( (2y - 3)(y - 2) = 0 \)
This gives \( y = \frac{3}{2} \) or \( y = 2 \).
The circle passes through R(0, 2), so the other point where the y-axis cuts the circle is \( (0, \frac{3}{2}) \).

**Part (ii): Coordinates of the other end of the diameter through Q**
First, find the centre of the circle from the values of \( g \) and \( f \).
Centre \( C = (-g, -f) = (-(-2), -(-\frac{7}{4})) = (2, \frac{7}{4}) \).
The diameter passes through Q(3, 0) and the centre \( C(2, \frac{7}{4}) \). Let the other end of the diameter be \( Q'(\alpha, \beta) \).
The centre C is the midpoint of the diameter \( QQ' \).
Using the midpoint formula:
\( \frac{x_1 + x_2}{2} = x_C \implies \frac{3 + \alpha}{2} = 2 \implies 3 + \alpha = 4 \implies \alpha = 1 \)
\( \frac{y_1 + y_2}{2} = y_C \implies \frac{0 + \beta}{2} = \frac{7}{4} \implies \beta = \frac{7}{2} \)
So, the coordinates of the other end of the diameter through Q are \( (1, \frac{7}{2}) \).
In simple words: First, we found the equation of the circle using the three given points. Then, we figured out where the circle crosses the y-axis again by setting \( x=0 \) in its equation. Finally, we found the centre of the circle and used the midpoint formula (since the centre is always the middle of a diameter) to find the other end of the diameter that passes through point Q. A diameter always passes through the center of a circle, making the center the midpoint of its endpoints.

🎯 Exam Tip: For multi-part questions, break them down and solve each part systematically. Remember that the centre of the circle is the midpoint of any of its diameters.

Question 16. Find the equation of the circle which has its centre on the line \( y = 2 \) and which passes through the points (2, 0) and (4, 0).
Answer:
Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
The centre of the circle is \( (-g, -f) \).
We are given that the centre lies on the line \( y = 2 \). This means the y-coordinate of the centre is 2.
So, \( -f = 2 \implies f = -2 \).

The circle passes through the point \( (2, 0) \). Substitute \( x=2, y=0 \) and \( f=-2 \) into the general equation:
\( 2^2 + 0^2 + 2g(2) + 2(-2)(0) + c = 0 \)
\( 4 + 4g + 0 + c = 0 \implies 4g + c + 4 = 0 \) (Equation 2)

The circle passes through the point \( (4, 0) \). Substitute \( x=4, y=0 \) and \( f=-2 \) into the general equation:
\( 4^2 + 0^2 + 2g(4) + 2(-2)(0) + c = 0 \)
\( 16 + 8g + 0 + c = 0 \implies 8g + c + 16 = 0 \) (Equation 3)

Now, we have two equations (Equation 2 and Equation 3) with two unknowns (\( g \) and \( c \)).
Subtract Equation 2 from Equation 3:
\( (8g + c + 16) - (4g + c + 4) = 0 \)
\( 4g + 12 = 0 \implies 4g = -12 \implies g = -3 \).

Substitute \( g=-3 \) into Equation 2:
\( 4(-3) + c + 4 = 0 \)
\( -12 + c + 4 = 0 \implies c - 8 = 0 \implies c = 8 \).

So, we have \( g = -3, f = -2, \) and \( c = 8 \).
Substitute these values back into the general equation of the circle:
\( x^2 + y^2 + 2(-3)x + 2(-2)y + 8 = 0 \)
\( x^2 + y^2 - 6x - 4y + 8 = 0 \).
This is the required equation of the circle.
In simple words: We used the general equation for a circle. The special hint that the centre lies on the line \( y=2 \) immediately told us the 'f' value. Then, by using the two points the circle passes through, we formed two more equations. Solving these equations let us find the 'g' and 'c' values, and then we wrote the full circle equation. Knowing that the center lies on a horizontal line (like y=2) directly gives you the y-coordinate of the center.

🎯 Exam Tip: If the centre lies on a given line, substitute the coordinates of the centre \( (-g, -f) \) into the line's equation to get a useful relationship between \( g \) and \( f \).

Question 17. Find the equation of the circle which passes through the points (1, -2),(4, -3) and has its centre on the line \( 3x + 4y + 10 = 0 \).
Answer:
Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

**1. Use the condition that the centre lies on the given line:**
The centre of the circle is \( (-g, -f) \).
It lies on the line \( 3x + 4y + 10 = 0 \). Substitute \( x=-g \) and \( y=-f \):
\( 3(-g) + 4(-f) + 10 = 0 \)
\( -3g - 4f + 10 = 0 \)
\( 3g + 4f - 10 = 0 \) (Equation 4)

**2. Use the points the circle passes through:**
For point \( (1, -2) \):
\( 1^2 + (-2)^2 + 2g(1) + 2f(-2) + c = 0 \)
\( 1 + 4 + 2g - 4f + c = 0 \)
\( 2g - 4f + c + 5 = 0 \) (Equation 2)

For point \( (4, -3) \):
\( 4^2 + (-3)^2 + 2g(4) + 2f(-3) + c = 0 \)
\( 16 + 9 + 8g - 6f + c = 0 \)
\( 8g - 6f + c + 25 = 0 \) (Equation 3)

**3. Solve the system of equations:**
Subtract Equation 2 from Equation 3:
\( (8g - 6f + c + 25) - (2g - 4f + c + 5) = 0 \)
\( 6g - 2f + 20 = 0 \)
Divide by 2: \( 3g - f + 10 = 0 \) (Equation 5)

Now we have a system of two equations (Equation 4 and Equation 5) with two unknowns (\( g \) and \( f \)):
\( 3g + 4f - 10 = 0 \) (Equation 4)
\( 3g - f + 10 = 0 \) (Equation 5)

From Equation 5, express \( f \) in terms of \( g \):
\( f = 3g + 10 \)
Substitute this expression for \( f \) into Equation 4:
\( 3g + 4(3g + 10) - 10 = 0 \)
\( 3g + 12g + 40 - 10 = 0 \)
\( 15g + 30 = 0 \)
\( 15g = -30 \implies g = -2 \).

Now find \( f \):
\( f = 3(-2) + 10 = -6 + 10 = 4 \).

Finally, find \( c \) using Equation 2:
\( 2g - 4f + c + 5 = 0 \)
\( 2(-2) - 4(4) + c + 5 = 0 \)
\( -4 - 16 + c + 5 = 0 \)
\( -15 + c = 0 \implies c = 15 \).

So, we have \( g = -2, f = 4, \) and \( c = 15 \).
Substitute these values back into the general equation of the circle:
\( x^2 + y^2 + 2(-2)x + 2(4)y + 15 = 0 \)
\( x^2 + y^2 - 4x + 8y + 15 = 0 \).
This is the required equation of the circle.
In simple words: We used the general circle equation. First, we wrote an equation for the centre point because it lies on a specific line. Then, we used the two points the circle passes through to get two more equations. We combined all these equations to find the special numbers (g, f, c) that make up the circle's equation. The condition that the circle's center lies on a specific line provides a crucial linear equation that helps solve for the unknown coefficients (g, f, c).

🎯 Exam Tip: When the centre lies on a line, substitute \( (-g, -f) \) into the line's equation to create one of the linear equations needed to solve for \( g, f, c \).

Question 18. The vertices A, B, C of a triangle ABC have co-ordinates (4, 4),(5, 3) and (6, 0) respectively. Find the equations of the perpendicular bisectors of AB and BC, the co-ordinates of the circumcentre and the radius of the circumcircle of the triangle ABC.
Answer: Let D and E be the mid-points of sides AB and BC of triangle ABC.
The coordinates of midpoint D (for AB) are \( \left(\frac{4+5}{2}, \frac{4+3}{2}\right) = \left(\frac{9}{2}, \frac{7}{2}\right) \).
The coordinates of midpoint E (for BC) are \( \left(\frac{5+6}{2}, \frac{3+0}{2}\right) = \left(\frac{11}{2}, \frac{3}{2}\right) \).
The slope of side AB \( = \frac{4-3}{4-5} = \frac{1}{-1} = -1 \).
Since OD is the perpendicular bisector of AB, its slope is the negative reciprocal of the slope of AB.
So, the slope of line OD \( = \frac{-1}{-1} = 1 \).
The equation of the perpendicular bisector OD, passing through \( \left(\frac{9}{2}, \frac{7}{2}\right) \) with slope 1, is given by:
\( y - \frac{7}{2} = 1 \left(x - \frac{9}{2}\right) \)
\( 2y - 7 = 2x - 9 \)
\( 2x - 2y - 2 = 0 \)
\( \implies \) \( x - y - 1 = 0 \) ...(1)
Now, the slope of side BC \( = \frac{0-3}{6-5} = \frac{-3}{1} = -3 \).
Since OE is the perpendicular bisector of BC, its slope is the negative reciprocal of the slope of BC.
So, the slope of line OE \( = \frac{-1}{-3} = \frac{1}{3} \).
The equation of the perpendicular bisector OE, passing through \( \left(\frac{11}{2}, \frac{3}{2}\right) \) with slope \( \frac{1}{3} \), is given by:
\( y - \frac{3}{2} = \frac{1}{3} \left(x - \frac{11}{2}\right) \)
\( 3(2y - 3) = 1(2x - 11) \)
\( 6y - 9 = 2x - 11 \)
\( 2x - 6y - 2 = 0 \)
\( \implies \) \( x - 3y - 1 = 0 \) ...(2)
The circumcentre is the point where these two perpendicular bisectors intersect. We solve equations (1) and (2).
From (1): \( x = y + 1 \). Substitute this into (2):
\( (y + 1) - 3y - 1 = 0 \)
\( -2y = 0 \)
\( \implies \) \( y = 0 \)
Substitute \( y = 0 \) back into \( x = y + 1 \):
\( x = 0 + 1 \)
\( \implies \) \( x = 1 \)
So, the circumcentre is (1, 0).
The radius of the circumcircle is the distance from the circumcentre (1, 0) to any of the vertices (A, B, or C). Let's use vertex B(5, 3).
Radius \( = \sqrt{(1-5)^2 + (0-3)^2} \)
\( = \sqrt{(-4)^2 + (-3)^2} \)
\( = \sqrt{16 + 9} \)
\( = \sqrt{25} = 5 \) units. The circumradius is the distance from the circumcenter to any vertex of the triangle.
In simple words: First, we find the middle points of two sides of the triangle. Then, we find the equations of lines that cut these sides in half at a right angle. Where these two lines cross is the center of the circle that goes through all three corners of the triangle. The distance from this center to any corner is the radius.

🎯 Exam Tip: Remember that the circumcentre is equidistant from all vertices of the triangle. Using the distance formula from the circumcentre to all three vertices A, B, and C can help verify your calculated radius.

Question 19. The radius of a circle is 5 units and it touches the circle x² + y² – 2x – 4y – 20 = 0 externally at the point (5, 5). Find the equation of the circle.
Answer: Let the given circle be \( S_1: x^2 + y^2 - 2x - 4y - 20 = 0 \).
Comparing this with the general equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \), we have \( 2g = -2 \implies g = -1 \) and \( 2f = -4 \implies f = -2 \). Also, \( c = -20 \).
The centre of circle \( S_1 \) is \( C_1 = (-g, -f) = (1, 2) \).
The radius of circle \( S_1 \) is \( r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-2)^2 - (-20)} = \sqrt{1 + 4 + 20} = \sqrt{25} = 5 \) units.
Let the required circle be \( S_2 \) with centre \( C_2 = (\alpha, \beta) \) and radius \( r_2 = 5 \) (given in the question).
The two circles touch externally at point P(5, 5). When two circles touch externally, the point of contact divides the line segment joining their centres in the ratio of their radii. Since \( r_1 = 5 \) and \( r_2 = 5 \), the ratio is 1:1. This means P(5, 5) is the midpoint of \( C_1C_2 \).
Using the midpoint formula for P(5, 5) and \( C_1(1, 2) \), we can find \( C_2(\alpha, \beta) \):
For x-coordinate: \( \frac{1 + \alpha}{2} = 5 \implies 1 + \alpha = 10 \implies \alpha = 9 \).
For y-coordinate: \( \frac{2 + \beta}{2} = 5 \implies 2 + \beta = 10 \implies \beta = 8 \).
So, the centre of the required circle \( S_2 \) is \( C_2(9, 8) \).
The equation of a circle with centre \((h, k)\) and radius \(r\) is \( (x-h)^2 + (y-k)^2 = r^2 \).
For circle \( S_2 \), \( h=9, k=8, r=5 \).
Equation of required circle is \( (x-9)^2 + (y-8)^2 = 5^2 \).
\( x^2 - 18x + 81 + y^2 - 16y + 64 = 25 \)
\( x^2 + y^2 - 18x - 16y + 145 = 25 \)
\( x^2 + y^2 - 18x - 16y + 120 = 0 \). The equation clearly shows its center and radius, which helps confirm its properties.
In simple words: We find the center and radius of the first circle. Since the two circles touch each other on the outside and have the same radius, the point where they touch is exactly halfway between their centers. We use this to find the center of our new circle. Then, with its center and radius, we write down its equation.

🎯 Exam Tip: When circles touch externally, the distance between their centers is the sum of their radii. For internal touch, it's the absolute difference of their radii. The point of contact always lies on the line connecting the two centers.

Question 20. Prove that the circle x² + y² – 2ax + 2y = 0 cuts the axis of x at the point A and it circumscribes an equilateral triangle with OA as one of the sides, where O is the origin. Find the value of a and the vertices of the equilateral triangle.
Answer: Let the given equation of the circle be \( x^2 + y^2 - 2ax + 2y = 0 \) ...(1)
To find where the circle meets the x-axis, we set \( y = 0 \) in equation (1):
\( x^2 - 2ax = 0 \)
\( x(x - 2a) = 0 \)
\( \implies \) \( x = 0 \) or \( x = 2a \).
This means the circle intersects the x-axis at O(0, 0) and A(2a, 0). This confirms the first part of the question.
Let's find the centre \( C \) and radius \( R \) of the circle (1).
Comparing with \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
\( 2g = -2a \implies g = -a \)
\( 2f = 2 \implies f = 1 \)
\( c = 0 \)
Centre \( C = (-g, -f) = (a, -1) \).
Radius \( R = \sqrt{g^2 + f^2 - c} = \sqrt{(-a)^2 + (1)^2 - 0} = \sqrt{a^2 + 1} \).
For an equilateral triangle OAC where O(0,0) and A(2a,0) are vertices, and OA is a side.
The side length of the triangle is \( s = OA = \sqrt{(2a-0)^2 + (0-0)^2} = \sqrt{4a^2} = 2a \).
For a circle to circumscribe an equilateral triangle, its radius \( R \) is related to the side length \( s \) by the formula \( R = \frac{s}{\sqrt{3}} \).
So, \( \sqrt{a^2 + 1} = \frac{2a}{\sqrt{3}} \).
Squaring both sides:
\( a^2 + 1 = \frac{4a^2}{3} \)
\( 3(a^2 + 1) = 4a^2 \)
\( 3a^2 + 3 = 4a^2 \)
\( a^2 = 3 \)
\( \implies \) \( a = \sqrt{3} \) (since A is on the positive x-axis, \( a > 0 \)).
Now we find the vertices of the equilateral triangle OAC.
O is the origin (0, 0).
A is \( (2a, 0) = (2\sqrt{3}, 0) \).
To find C, for an equilateral triangle OAC where OA is on the x-axis, the third vertex C can be found using rotation or properties of equilateral triangles. The x-coordinate of C is the midpoint of OA, which is \( \frac{0+2a}{2} = a \). The y-coordinate is \( \pm a\sqrt{3} \) (height of equilateral triangle of side \(2a\) is \( (2a)\frac{\sqrt{3}}{2} = a\sqrt{3} \)). Since the solution on the next page uses a negative y-coordinate for C, we will follow that.
So, \( C = (a, -a\sqrt{3}) = (\sqrt{3}, -\sqrt{3} \cdot \sqrt{3}) = (\sqrt{3}, -3) \).
Thus, the value of \( a = \sqrt{3} \), and the vertices of the equilateral triangle are \( O(0, 0) \), \( A(2\sqrt{3}, 0) \), and \( C(\sqrt{3}, -3) \). The circumradius of an equilateral triangle is the distance from the circumcenter to any vertex.
In simple words: First, we show that the circle crosses the x-axis at the starting point (origin) and another point A. Then, we use the rule that for a circle around a three-sided shape with equal sides (equilateral triangle), its size is linked to the length of the triangle's side. This helps us find the exact number for 'a' and the positions of all three corners of the triangle.

🎯 Exam Tip: For equilateral triangles, remember the relationship between the circumradius (R) and the side length (s): \( R = \frac{s}{\sqrt{3}} \). This formula is key to solving problems involving circumscribing circles for equilateral triangles.

Question 21. Find the equation of the circle which passes through the points (5, 0) and (1, 4) and whose centre lies on the line x + y – 3 = 0.
Answer: Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \) ...(1)
The circle passes through the point (5, 0). Substitute \( x=5, y=0 \) into (1):
\( 5^2 + 0^2 + 2g(5) + 2f(0) + c = 0 \)
\( 25 + 10g + c = 0 \) ...(2)
The circle passes through the point (1, 4). Substitute \( x=1, y=4 \) into (1):
\( 1^2 + 4^2 + 2g(1) + 2f(4) + c = 0 \)
\( 1 + 16 + 2g + 8f + c = 0 \)
\( 2g + 8f + c + 17 = 0 \) ...(3)
The centre of the circle is \( (-g, -f) \). This centre lies on the line \( x + y - 3 = 0 \).
So, substitute \( x=-g, y=-f \) into the line equation:
\( -g - f - 3 = 0 \)
\( \implies \) \( g + f + 3 = 0 \) ...(4)
Now, we solve the system of equations (2), (3), and (4) to find g, f, and c.
Subtract equation (3) from equation (2):
\( (25 + 10g + c) - (2g + 8f + c + 17) = 0 - 0 \)
\( 25 + 10g + c - 2g - 8f - c - 17 = 0 \)
\( 8g - 8f + 8 = 0 \)
Divide by 8: \( g - f + 1 = 0 \) ...(5)
Now we have a system of two equations for g and f: (4) and (5).
(4) \( g + f = -3 \)
(5) \( g - f = -1 \)
Add (4) and (5):
\( (g + f) + (g - f) = -3 + (-1) \)
\( 2g = -4 \)
\( \implies \) \( g = -2 \)
Substitute \( g = -2 \) into (4):
\( -2 + f = -3 \)
\( \implies \) \( f = -1 \)
Substitute \( g = -2 \) and \( f = -1 \) into equation (2):
\( 25 + 10(-2) + c = 0 \)
\( 25 - 20 + c = 0 \)
\( 5 + c = 0 \)
\( \implies \) \( c = -5 \)
Now, substitute the values of g, f, and c into the general equation of the circle (1):
\( x^2 + y^2 + 2(-2)x + 2(-1)y + (-5) = 0 \)
\( x^2 + y^2 - 4x - 2y - 5 = 0 \). This final equation represents the circle that fits all the given conditions.
In simple words: We start with the general math formula for a circle. Then, we use the two given points that the circle passes through to create two math problems. We also use the information that the center of the circle lies on a certain straight line to create a third problem. By solving these three math problems together, we find the specific numbers needed to write the exact equation of the circle.

🎯 Exam Tip: To find the equation of a circle, you typically need three independent conditions. These conditions are often points the circle passes through, tangency to a line, or information about the centre or radius. Formulate equations from each condition and solve simultaneously.