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

Exercise 17(A)

Question 1. Find the equation of the circle
(i) whose centre is (4, 5), radius is 7 ;
(ii) whose centre is (0, – 4) and which touches the x-axis ;
(iii) which passes through the origin and cuts off intercepts of length 'a', each from positive direction of the axes.
Answer:
(i) The equation of a circle with centre \( (h, k) \) and radius \( r \) is given by \( (x-h)^2 + (y-k)^2 = r^2 \).
Here, the centre is \( (4, 5) \) and the radius is \( 7 \).
So, the equation is \( (x-4)^2 + (y-5)^2 = 7^2 \).
Expanding this, we get \( x^2 - 8x + 16 + y^2 - 10y + 25 = 49 \).
Combining terms, the equation of the circle is \( x^2 + y^2 - 8x - 10y - 8 = 0 \). A circle is uniquely defined by its center and radius.
(ii) The centre of the circle is \( (0, -4) \). Since the circle touches the x-axis, its radius is the absolute value of the y-coordinate of the centre.
So, the radius \( r = |-4| = 4 \).
Using the general equation of a circle, \( (x-h)^2 + (y-k)^2 = r^2 \).
Here, \( h=0, k=-4, r=4 \).
\( (x-0)^2 + (y-(-4))^2 = 4^2 \).
\( x^2 + (y+4)^2 = 16 \).
Expanding this, we get \( x^2 + y^2 + 8y + 16 = 16 \).
The equation of the circle is \( x^2 + y^2 + 8y = 0 \). The radius is the distance from the center to any point on the circle.
(iii) The circle passes through the origin \( (0, 0) \) and cuts off intercepts of length 'a' from the positive directions of the axes. This means it passes through points \( (a, 0) \) on the x-axis and \( (0, a) \) on the y-axis.
These two points, \( (a, 0) \) and \( (0, a) \), form the ends of a diameter of the circle because the circle passes through the origin and cuts off equal intercepts 'a' from both positive axes.
The equation of a circle with diameter endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( (x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0 \).
Using \( (a, 0) \) and \( (0, a) \) as the diameter endpoints:
\( (x-a)(x-0) + (y-0)(y-a) = 0 \).
\( x(x-a) + y(y-a) = 0 \).
Expanding this gives \( x^2 - ax + y^2 - ay = 0 \).
Rearranging, the equation of the circle is \( x^2 + y^2 - ax - ay = 0 \). This form is useful when two points on the diameter are known.
In simple words: (i) To find the circle's equation, put the centre and radius into the standard formula. (ii) If a circle touches the x-axis, its radius is the y-coordinate of its centre. (iii) If a circle passes through the origin and cuts equal parts from both axes, those cut points can be thought of as the ends of its diameter.

🎯 Exam Tip: Remember the standard form of a circle's equation: \( (x-h)^2 + (y-k)^2 = r^2 \). For parts (ii) and (iii), visualize the geometry to correctly determine the radius or diameter endpoints.

Question 2. Find the equation of the circle
(i) whose centre is (a, b) and which passes through the origin ;
(ii) whose centre is the point (2, 3) and which passes through the intersection of the lines \( 3x - 2y - 1 = 0 \) and \( 4x + y - 27 = 0 \).
Answer:
(i) The centre of the circle is \( (a, b) \). The circle passes through the origin \( (0, 0) \).
The radius \( r \) of the circle is the distance between the centre \( (a, b) \) and the origin \( (0, 0) \).
\( r = \sqrt{(a-0)^2 + (b-0)^2} = \sqrt{a^2 + b^2} \).
So, \( r^2 = a^2 + b^2 \). The distance formula helps find the radius here.
Using the standard equation of a circle \( (x-h)^2 + (y-k)^2 = r^2 \), with \( h=a, k=b \):
\( (x-a)^2 + (y-b)^2 = a^2 + b^2 \).
Expanding this: \( x^2 - 2ax + a^2 + y^2 - 2by + b^2 = a^2 + b^2 \).
Subtracting \( a^2 + b^2 \) from both sides, the equation of the circle is \( x^2 + y^2 - 2ax - 2by = 0 \).
(ii) The centre of the circle is \( C(2, 3) \). First, we need to find the point of intersection of the two given lines.
Line (1): \( 3x - 2y - 1 = 0 \)
Line (2): \( 4x + y - 27 = 0 \)
Multiply equation (2) by 2:
\( 2(4x + y - 27) = 0 \implies 8x + 2y - 54 = 0 \) (Equation 3).
Add Equation (1) and Equation (3):
\( (3x - 2y - 1) + (8x + 2y - 54) = 0 \)
\( 11x - 55 = 0 \).
\( 11x = 55 \implies x = 5 \).
Substitute \( x=5 \) into Equation (2):
\( 4(5) + y - 27 = 0 \)
\( 20 + y - 27 = 0 \)
\( y - 7 = 0 \implies y = 7 \).
So, the point of intersection is \( P(5, 7) \). This point lies on the circle.
The radius \( r \) of the circle is the distance between the centre \( C(2, 3) \) and the point \( P(5, 7) \).
\( r = \sqrt{(5-2)^2 + (7-3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5 \).
So, \( r^2 = 25 \).
Using the standard equation of a circle \( (x-h)^2 + (y-k)^2 = r^2 \), with \( h=2, k=3 \):
\( (x-2)^2 + (y-3)^2 = 5^2 \).
\( (x-2)^2 + (y-3)^2 = 25 \).
Expanding this: \( x^2 - 4x + 4 + y^2 - 6y + 9 = 25 \).
\( x^2 + y^2 - 4x - 6y + 13 = 25 \).
The equation of the circle is \( x^2 + y^2 - 4x - 6y - 12 = 0 \). Solving simultaneous equations is key to finding the intersection point.
In simple words: (i) If you know the centre and a point the circle passes through (like the origin), the distance between them is the radius. Use this to write the circle's equation. (ii) First, find where the two lines cross by solving their equations. This crossing point is on the circle. Then, use this point and the given centre to find the radius, and finally, the circle's equation.

🎯 Exam Tip: When finding the radius, ensure you use the distance formula correctly. For intersecting lines, careful algebraic manipulation is needed to find the exact point.

Question 3. Find the equation of the circle which has A(1, 3) and B(4, 5) as opposite ends of a diameter. Find also the equation of the perpendicular diameter.
Answer:
For a circle where \( A(1, 3) \) and \( B(4, 5) \) are the endpoints of a diameter, the equation can be found using the diametrical form:
\( (x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0 \).
Here, \( x_1=1, y_1=3, x_2=4, y_2=5 \).
\( (x-1)(x-4) + (y-3)(y-5) = 0 \).
Expand the terms:
\( (x^2 - 4x - x + 4) + (y^2 - 5y - 3y + 15) = 0 \).
\( x^2 - 5x + 4 + y^2 - 8y + 15 = 0 \).
The equation of the circle is \( x^2 + y^2 - 5x - 8y + 19 = 0 \). This formula simplifies finding the circle's equation when two points on its diameter are known.

Next, we find the equation of the perpendicular diameter.
The centre of the circle C is the midpoint of the diameter AB.
Coordinates of C \( = \left( \frac{1+4}{2}, \frac{3+5}{2} \right) = \left( \frac{5}{2}, \frac{8}{2} \right) = \left( \frac{5}{2}, 4 \right) \).
The slope of the diameter AB is \( m_{AB} = \frac{y_2-y_1}{x_2-x_1} = \frac{5-3}{4-1} = \frac{2}{3} \).
A diameter perpendicular to AB will have a slope that is the negative reciprocal of \( m_{AB} \).
Slope of perpendicular diameter \( m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2} \).
The perpendicular diameter also passes through the centre \( C\left( \frac{5}{2}, 4 \right) \).
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 4 = -\frac{3}{2} \left( x - \frac{5}{2} \right) \).
Multiply both sides by 2:
\( 2(y-4) = -3 \left( x - \frac{5}{2} \right) \).
\( 2y - 8 = -3x + \frac{15}{2} \).
Multiply by 2 again to clear the fraction:
\( 2(2y - 8) = 2\left( -3x + \frac{15}{2} \right) \).
\( 4y - 16 = -6x + 15 \).
Rearrange into the general form \( Ax + By + C = 0 \):
\( 6x + 4y - 16 - 15 = 0 \).
The equation of the perpendicular diameter is \( 6x + 4y - 31 = 0 \). The negative reciprocal of a slope gives the perpendicular slope.
In simple words: To find the circle's equation when you have the ends of its diameter, use a special formula. For the perpendicular diameter, first find the middle point of the diameter (which is the circle's centre). Then, find the slope of the first diameter. The perpendicular diameter will have the opposite, flipped slope, and it will also pass through the centre.

🎯 Exam Tip: Remember that the product of slopes of perpendicular lines is -1. Also, every diameter passes through the circle's center, which is the midpoint of any diameter.

Question 4. Find the equation to the circles which pass through the origin and cut off intercepts equal to (i) 2 and 4, (ii) 2a and 2b from the x-axis and the y-axis respectively.
Answer:
(i) The circle passes through the origin \( (0, 0) \) and cuts off intercepts of length 2 from the x-axis and 4 from the y-axis. This means the circle passes through points \( (2, 0) \) and \( (0, 4) \). In the given solution, it seems to imply that the intercepts are \( (3,0) \) and \( (0,4) \). Following the solution's calculation, let's assume the points are \( (3,0) \) and \( (0,4) \) which are the endpoints of a diameter.
The equation of a circle with diameter endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( (x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0 \).
Using \( (3, 0) \) and \( (0, 4) \) as the diameter endpoints:
\( (x-3)(x-0) + (y-0)(y-4) = 0 \).
\( x(x-3) + y(y-4) = 0 \).
Expanding this gives \( x^2 - 3x + y^2 - 4y = 0 \).
Rearranging, the equation of the circle is \( x^2 + y^2 - 3x - 4y = 0 \). This method is useful when the x and y-intercepts define the diameter.
(ii) The circle passes through the origin \( (0, 0) \) and cuts off intercepts of length \( 2a \) from the x-axis and \( 2b \) from the y-axis. This implies the circle passes through points \( (2a, 0) \) and \( (0, 2b) \). These two points are the endpoints of a diameter.
Using the diametrical form of the circle's equation with endpoints \( (2a, 0) \) and \( (0, 2b) \):
\( (x-2a)(x-0) + (y-0)(y-2b) = 0 \).
\( x(x-2a) + y(y-2b) = 0 \).
Expanding this gives \( x^2 - 2ax + y^2 - 2by = 0 \).
Rearranging, the equation of the circle is \( x^2 + y^2 - 2ax - 2by = 0 \). This applies when the circle passes through the origin and its intercepts form the diameter.
In simple words: When a circle goes through the origin and cuts off parts on the x and y axes, those cut-off points can be used as the ends of its diameter to write down the circle's equation.

🎯 Exam Tip: If a circle passes through the origin and makes intercepts on the axes, these intercepts can be considered as the coordinates of the endpoints of a diameter. Remember that the x-intercept will have a y-coordinate of 0, and the y-intercept will have an x-coordinate of 0.

Question 5. Find the equation of the circles which touch the axis of x at a distance of 4 from the origin and cut off an intercept of 6 from the axis of y.
Answer:
The circle touches the x-axis at a distance of 4 from the origin. This means the point of tangency on the x-axis is \( (4, 0) \) or \( (-4, 0) \).
If the circle touches the x-axis at \( (4, 0) \), the centre of the circle must have an x-coordinate of 4. Let the centre be \( (4, k) \). The radius of the circle will be \( r = |k| \).
The equation of such a circle is \( (x-4)^2 + (y-k)^2 = k^2 \).
The circle cuts off an intercept of length 6 from the y-axis. This means when \( x=0 \), the difference between the two y-values where the circle crosses the y-axis is 6.
Substitute \( x=0 \) into the equation:
\( (0-4)^2 + (y-k)^2 = k^2 \)
\( 16 + y^2 - 2yk + k^2 = k^2 \).
\( y^2 - 2yk + 16 = 0 \).
Let the roots of this quadratic equation be \( y_1 \) and \( y_2 \). The length of the y-intercept is \( |y_1 - y_2| = 6 \).
For a quadratic equation \( Ay^2 + By + C = 0 \), \( |y_1 - y_2| = \frac{\sqrt{B^2 - 4AC}}{|A|} \).
Here, \( A=1, B=-2k, C=16 \).
\( 6 = \frac{\sqrt{(-2k)^2 - 4(1)(16)}}{1} \).
\( 6 = \sqrt{4k^2 - 64} \).
Square both sides: \( 36 = 4k^2 - 64 \).
\( 4k^2 = 36 + 64 \).
\( 4k^2 = 100 \).
\( k^2 = 25 \implies k = \pm 5 \).
So, the possible centres are \( (4, 5) \) and \( (4, -5) \). The radius for both is \( r = |k| = 5 \). This calculation gives us the possible y-coordinates for the center.

Equations of the circles for x-tangency at \( (4,0) \):
1. Centre \( (4, 5) \), radius 5:
\( (x-4)^2 + (y-5)^2 = 5^2 \)
\( x^2 - 8x + 16 + y^2 - 10y + 25 = 25 \)
\( x^2 + y^2 - 8x - 10y + 16 = 0 \).
2. Centre \( (4, -5) \), radius 5:
\( (x-4)^2 + (y-(-5))^2 = 5^2 \)
\( (x-4)^2 + (y+5)^2 = 25 \)
\( x^2 - 8x + 16 + y^2 + 10y + 25 = 25 \)
\( x^2 + y^2 - 8x + 10y + 16 = 0 \).

Similarly, if the circle touches the x-axis at \( (-4, 0) \), the centre of the circle must have an x-coordinate of -4. Let the centre be \( (-4, k) \). The radius will be \( r = |k| \).
The equation of such a circle is \( (x-(-4))^2 + (y-k)^2 = k^2 \).
\( (x+4)^2 + (y-k)^2 = k^2 \).
Using the same logic for the y-intercept, we will again find \( k = \pm 5 \).
This gives two more possible centres: \( (-4, 5) \) and \( (-4, -5) \). The radius for both is \( r = |k| = 5 \).

Equations of the circles for x-tangency at \( (-4,0) \):
3. Centre \( (-4, 5) \), radius 5:
\( (x-(-4))^2 + (y-5)^2 = 5^2 \)
\( (x+4)^2 + (y-5)^2 = 25 \)
\( x^2 + 8x + 16 + y^2 - 10y + 25 = 25 \)
\( x^2 + y^2 + 8x - 10y + 16 = 0 \).
4. Centre \( (-4, -5) \), radius 5:
\( (x-(-4))^2 + (y-(-5))^2 = 5^2 \)
\( (x+4)^2 + (y+5)^2 = 25 \)
\( x^2 + 8x + 16 + y^2 + 10y + 25 = 25 \)
\( x^2 + y^2 + 8x + 10y + 16 = 0 \).
Therefore, there are four such circles. The symmetry of the problem creates these multiple solutions.
In simple words: First, understand that touching the x-axis at a distance of 4 from the origin means the x-coordinate of the centre is 4 or -4, and the y-coordinate gives the radius. Then, use the information about the y-intercept to find the radius (y-coordinate). Combine these to find all four possible circles.

🎯 Exam Tip: Remember that "distance from the origin" on an axis can be positive or negative, leading to multiple possible tangent points. When a circle touches an axis, its radius is equal to the absolute value of the coordinate perpendicular to that axis (e.g., radius = \( |k| \) for x-axis tangency). Pay attention to all possible quadrants.

Question 6. A circle having its centre in the first quadrant touches the y-axis at the point (0, 2) and passes through the point (1, 0). Find the equation of the circle.
Answer:
The circle touches the y-axis at the point \( (0, 2) \). This means the y-coordinate of the centre is 2. Let the centre of the circle be \( (h, 2) \).
Since the circle touches the y-axis, its radius \( r \) is the absolute value of the x-coordinate of its centre. So, \( r = |h| \).
The problem states the centre is in the first quadrant, so \( h \) must be positive, which means \( r = h \).
The equation of the circle is \( (x-h)^2 + (y-2)^2 = h^2 \). The x-coordinate of the center becomes the radius.
The circle also passes through the point \( (1, 0) \). Substitute these coordinates into the equation:
\( (1-h)^2 + (0-2)^2 = h^2 \).
\( 1 - 2h + h^2 + (-2)^2 = h^2 \).
\( 1 - 2h + h^2 + 4 = h^2 \).
\( 5 - 2h = 0 \).
\( 2h = 5 \implies h = \frac{5}{2} \).
So, the centre of the circle is \( \left( \frac{5}{2}, 2 \right) \) and the radius \( r = \frac{5}{2} \).
Now substitute these values back into the standard equation of a circle:
\( \left( x - \frac{5}{2} \right)^2 + (y-2)^2 = \left( \frac{5}{2} \right)^2 \).
Expand the terms:
\( x^2 - 2 \cdot x \cdot \frac{5}{2} + \left(\frac{5}{2}\right)^2 + y^2 - 4y + 4 = \left(\frac{5}{2}\right)^2 \).
\( x^2 - 5x + \frac{25}{4} + y^2 - 4y + 4 = \frac{25}{4} \).
Subtract \( \frac{25}{4} \) from both sides:
\( x^2 + y^2 - 5x - 4y + 4 = 0 \). This is the final equation of the circle. Knowing the quadrant of the center helps determine the sign of the radius.
In simple words: When a circle touches the y-axis, its radius is the x-coordinate of its centre. Since it's in the first quarter of the graph, this x-coordinate is positive. Use this fact, along with the point the circle passes through, to find the exact centre and radius, then write its equation.

🎯 Exam Tip: Pay close attention to keywords like "touches the y-axis" and "first quadrant" as they provide crucial information about the coordinates of the centre and the sign of the radius. If the problem states "touches x-axis", the radius will be the y-coordinate of the center.

Question 7. Find the equation of the circle, radius 2 units, which lies in the positive quadrant and touches both axes of coordinates. Find also the equation of the circle with centre (6, 5) which touches the above circle externally.
Answer:
First part: Find the equation of the circle with radius 2 units, lying in the positive quadrant and touching both axes.
If a circle lies in the positive quadrant and touches both the x-axis and the y-axis, its centre must be \( (r, r) \), where \( r \) is the radius.
Given that the radius is 2 units, the centre of this circle (let's call it Circle 1) is \( C_1(2, 2) \).
The equation of Circle 1 is \( (x-2)^2 + (y-2)^2 = 2^2 \).
Expanding this: \( x^2 - 4x + 4 + y^2 - 4y + 4 = 4 \).
The equation of Circle 1 is \( x^2 + y^2 - 4x - 4y + 4 = 0 \).

Second part: Find the equation of the circle with centre \( C_2(6, 5) \) which touches Circle 1 externally.
Let the radius of this second circle (Circle 2) be \( r_2 \).
For two circles to touch externally, the distance between their centres must be equal to the sum of their radii.
Distance \( C_1C_2 = r_1 + r_2 \).
Here, \( C_1(2, 2) \), \( r_1=2 \), \( C_2(6, 5) \).
\( \text{Distance } C_1C_2 = \sqrt{(6-2)^2 + (5-2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5 \).
So, \( 5 = r_1 + r_2 \).
\( 5 = 2 + r_2 \).
\( r_2 = 3 \).
Now we have the centre \( C_2(6, 5) \) and radius \( r_2=3 \) for Circle 2.
The equation of Circle 2 is \( (x-6)^2 + (y-5)^2 = 3^2 \).
Expanding this: \( x^2 - 12x + 36 + y^2 - 10y + 25 = 9 \).
\( x^2 + y^2 - 12x - 10y + 61 = 9 \).
The equation of Circle 2 is \( x^2 + y^2 - 12x - 10y + 52 = 0 \). The sum of radii equals the distance between centers for externally touching circles.
In simple words: For the first circle, if it touches both axes in the positive part of the graph and has a radius of 2, its centre is at (2,2). Write its equation. For the second circle, you know its centre. Since it touches the first circle from the outside, the distance between their centres is equal to the sum of their radii. Use this to find the second circle's radius and then its equation.

🎯 Exam Tip: Remember the condition for two circles to touch externally: the distance between their centers is equal to the sum of their radii. Also, for a circle touching both axes, its center coordinates are \( (r, r) \).

Question 8. Obtain the equation of the circle, centre (1, 0), which passes through the point \( \mathrm{P}\left(3, \frac{3}{2}\right) \). Find also the equation of the equal circle which touches the given circle externally at P.
Answer:
First part: Equation of the circle with centre \( C(1, 0) \) passing through point \( P\left(3, \frac{3}{2}\right) \).
The radius of this circle (let's call it Circle 1) is the distance between its centre C and the point P.
\( r_1 = CP = \sqrt{(3-1)^2 + \left(\frac{3}{2}-0\right)^2} \).
\( r_1 = \sqrt{2^2 + \left(\frac{3}{2}\right)^2} = \sqrt{4 + \frac{9}{4}} \).
\( r_1 = \sqrt{\frac{16}{4} + \frac{9}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2} \).
So, \( r_1^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \).
The equation of Circle 1 is \( (x-1)^2 + (y-0)^2 = \left(\frac{5}{2}\right)^2 \).
\( (x-1)^2 + y^2 = \frac{25}{4} \).
Expand and multiply by 4 to clear the fraction:
\( 4(x^2 - 2x + 1) + 4y^2 = 25 \).
\( 4x^2 - 8x + 4 + 4y^2 = 25 \).
The equation of Circle 1 is \( 4x^2 + 4y^2 - 8x - 21 = 0 \). The distance formula helps determine the radius.

Second part: Equation of an equal circle that touches Circle 1 externally at point P.
An "equal circle" means it has the same radius as Circle 1, so \( r_2 = r_1 = \frac{5}{2} \).
This new circle (Circle 2) touches Circle 1 externally at point \( P\left(3, \frac{3}{2}\right) \).
When two circles touch externally, the point of contact P lies on the line segment joining their centres \( C \) and \( C' \). Since the radii are equal \( (r_1 = r_2) \), P must be the midpoint of the segment \( CC' \).
Let the centre of Circle 2 be \( C'(h, k) \).
We have \( C(1, 0) \) and \( P\left(3, \frac{3}{2}\right) \).
Using the midpoint formula:
\( \frac{1+h}{2} = 3 \implies 1+h = 6 \implies h = 5 \).
\( \frac{0+k}{2} = \frac{3}{2} \implies k = 3 \).
So, the centre of Circle 2 is \( C'(5, 3) \), and its radius is \( r_2 = \frac{5}{2} \).
The equation of Circle 2 is \( (x-5)^2 + (y-3)^2 = \left(\frac{5}{2}\right)^2 \).
\( (x-5)^2 + (y-3)^2 = \frac{25}{4} \).
Expand and multiply by 4 to clear the fraction:
\( 4(x^2 - 10x + 25) + 4(y^2 - 6y + 9) = 25 \).
\( 4x^2 - 40x + 100 + 4y^2 - 24y + 36 = 25 \).
\( 4x^2 + 4y^2 - 40x - 24y + 136 = 25 \).
The equation of Circle 2 is \( 4x^2 + 4y^2 - 40x - 24y + 111 = 0 \). The midpoint formula is crucial for finding the center of the touching circle.
In simple words: First, use the given centre and the point the circle passes through to find its radius. Then write its equation. For the second circle, since it's the same size and touches the first circle at a specific point, that point must be exactly in the middle of the two circles' centres. Use this to find the second circle's centre, then write its equation.

🎯 Exam Tip: When two circles of equal radii touch externally, the point of tangency is the midpoint of the line segment connecting their centers. Utilize the midpoint formula effectively in such scenarios.

Question 9. Calculate the co-ordinates of the foot of the perpendicular from the point (- 4, 2) to the line \( 3x + 2y = 5 \). Find the equation of the smallest circle passing through (- 4, 2) and having its centre on the line \( 3x + 2y = 5 \).
Answer:
First part: Coordinates of the foot of the perpendicular from \( P(-4, 2) \) to the line \( L_1: 3x + 2y = 5 \).
The slope of line \( L_1 \) is \( m_1 = -\frac{\text{coefficient of } x}{\text{coefficient of } y} = -\frac{3}{2} \).
The line perpendicular to \( L_1 \) (let's call it PM) will have a slope \( m_2 \) such that \( m_1 m_2 = -1 \).
\( m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3} \).
The line PM passes through \( P(-4, 2) \) and has a slope of \( \frac{2}{3} \). Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 2 = \frac{2}{3}(x - (-4)) \).
\( y - 2 = \frac{2}{3}(x+4) \).
Multiply by 3: \( 3(y-2) = 2(x+4) \).
\( 3y - 6 = 2x + 8 \).
Rearrange to form line \( L_2: 2x - 3y + 14 = 0 \). The foot of the perpendicular M is the intersection of \( L_1 \) and \( L_2 \).
We have two equations:
1) \( 3x + 2y = 5 \)
2) \( 2x - 3y = -14 \)
Multiply Equation (1) by 3: \( 9x + 6y = 15 \).
Multiply Equation (2) by 2: \( 4x - 6y = -28 \).
Add the two new equations:
\( (9x + 6y) + (4x - 6y) = 15 + (-28) \).
\( 13x = -13 \implies x = -1 \).
Substitute \( x=-1 \) into \( 3x + 2y = 5 \):
\( 3(-1) + 2y = 5 \).
\( -3 + 2y = 5 \).
\( 2y = 8 \implies y = 4 \).
So, the coordinates of the foot of the perpendicular M are \( (-1, 4) \). This is a point of perpendicular intersection.

Second part: Equation of the smallest circle passing through \( (-4, 2) \) and having its centre on the line \( 3x + 2y = 5 \).
The smallest circle that passes through a given point \( P \) and has its centre on a given line \( L \) will have its centre at the foot of the perpendicular from \( P \) to \( L \).
Therefore, the centre of this smallest circle is \( C = M(-1, 4) \).
The radius of this circle will be the distance from its centre \( M(-1, 4) \) to the point it passes through, \( P(-4, 2) \).
\( r = MP = \sqrt{(-4 - (-1))^2 + (2 - 4)^2} \).
\( r = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \).
So, \( r^2 = 13 \).
The equation of the circle is \( (x-h)^2 + (y-k)^2 = r^2 \), with \( h=-1, k=4 \).
\( (x-(-1))^2 + (y-4)^2 = (\sqrt{13})^2 \).
\( (x+1)^2 + (y-4)^2 = 13 \).
Expanding this: \( x^2 + 2x + 1 + y^2 - 8y + 16 = 13 \).
\( x^2 + y^2 + 2x - 8y + 17 = 13 \).
The equation of the smallest circle is \( x^2 + y^2 + 2x - 8y + 4 = 0 \). The shortest distance from a point to a line is along the perpendicular.
In simple words: First, find the line that goes through point P and is exactly straight up-and-down (perpendicular) to the main line. Where these two lines cross is the 'foot of the perpendicular'. This point is also the centre of the smallest circle that passes through P and has its centre on the main line. The distance from this centre to P is the radius of that circle.

🎯 Exam Tip: Remember that the foot of the perpendicular from a point to a line is the closest point on the line to the given point. This shortest distance forms the radius of the smallest circle with its center on the line and passing through the point.

Question 10. The point diametrically opposite to the point P(1, 0) on the circle \( x^2 + y^2 + 2x + 4y - 3 = 0 \) is
(a) (3, – 4)
(b) (- 3, 4)
(c) (-3, – 4)
(d) (3, 4)
Answer: (c) (-3, – 4)
To find the point diametrically opposite to \( P(1, 0) \), we can use the fact that \( P \) and its diametrically opposite point \( Q \) will have the center of the circle as their midpoint. First, find the center of the given circle.
The general equation of a circle is \( x^2 + y^2 + 2gx + 2fy + c = 0 \). The centre of the circle is \( (-g, -f) \).
For the given circle \( x^2 + y^2 + 2x + 4y - 3 = 0 \):
\( 2g = 2 \implies g = 1 \).
\( 2f = 4 \implies f = 2 \).
So, the centre of the circle is \( C(-1, -2) \).
Let \( Q(x_2, y_2) \) be the point diametrically opposite to \( P(1, 0) \).
Since \( C \) is the midpoint of \( PQ \):
\( \frac{1+x_2}{2} = -1 \implies 1+x_2 = -2 \implies x_2 = -3 \).
\( \frac{0+y_2}{2} = -2 \implies y_2 = -4 \).
Thus, the point diametrically opposite to \( P(1, 0) \) is \( (-3, -4) \). The center of a circle is always the midpoint of its diameter.
You can also check which of the options lies on the circle and forms a diameter with P(1,0).
For option (c) \( (-3, -4) \), substitute into the circle equation:
\( (-3)^2 + (-4)^2 + 2(-3) + 4(-4) - 3 \)
\( = 9 + 16 - 6 - 16 - 3 \)
\( = 25 - 25 = 0 \).
Since the value is 0, the point \( (-3, -4) \) lies on the circle. Also, the midpoint of \( (1,0) \) and \( (-3,-4) \) is \( (\frac{1-3}{2}, \frac{0-4}{2}) = (-1,-2) \), which is the centre of the circle. This confirms that it is the diametrically opposite point.
In simple words: First, find the centre of the circle from its equation. Since P and the opposite point are at the ends of a diameter, the centre must be exactly in the middle of them. Use this to find the coordinates of the opposite point.

🎯 Exam Tip: To find the point diametrically opposite to a given point on a circle, first identify the center of the circle from its equation. Then, use the midpoint formula, as the center is the midpoint of any diameter.