OP Malhotra Class 11 Maths Solutions Chapter 17 Circle Chapter Test

Question 1. Find the centre and radius of the circle \( 2x^2 + 2y^2 - x = 0 \).
Answer:
Let the given equation of the circle be \( 2x^2 + 2y^2 - x = 0 \).
To find the centre and radius, we first need to make the coefficients of \( x^2 \) and \( y^2 \) equal to 1. We do this by dividing the entire equation by 2.
\( \implies x^2 + y^2 - \frac{x}{2} = 0 \)
Now, we compare this equation with the standard general form of a circle, which is \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
By comparing the coefficients, we get:
\( 2g = -\frac{1}{2} \implies g = -\frac{1}{4} \)
\( 2f = 0 \implies f = 0 \)
\( c = 0 \)
The coordinates of the centre of the circle are given by \( (-g, -f) \).
\( \implies \) Centre \( = \left( -(-\frac{1}{4}), -0 \right) = \left( \frac{1}{4}, 0 \right) \)
The radius of the circle is calculated using the formula \( r = \sqrt{g^2 + f^2 - c} \).
\( \implies r = \sqrt{\left(-\frac{1}{4}\right)^2 + (0)^2 - 0} \)
\( \implies r = \sqrt{\frac{1}{16} + 0 - 0} \)
\( \implies r = \sqrt{\frac{1}{16}} = \frac{1}{4} \)
So, the centre of the circle is \( \left( \frac{1}{4}, 0 \right) \) and its radius is \( \frac{1}{4} \).
In simple words: To find the centre and radius, first divide the equation so \( x^2 \) and \( y^2 \) have a 1 in front. Then, use the standard circle formula to find the centre coordinates and calculate the radius.

🎯 Exam Tip: Always ensure the coefficients of \( x^2 \) and \( y^2 \) are 1 before identifying g, f, and c. Pay close attention to the signs when finding the centre coordinates.

Question 2. Find the equation of the circle with centre \( (-a, -b) \) and radius \( \sqrt{a^2-b^2} \).
Answer:
We are given the centre of the circle as \( (-a, -b) \) and the radius as \( \sqrt{a^2-b^2} \).
The standard centre-radius form of a circle's equation is \( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h, k) \) is the centre and \( r \) is the radius.
Here, \( h = -a \), \( k = -b \), and \( r = \sqrt{a^2-b^2} \).
Substitute these values into the formula:
\( (x - (-a))^2 + (y - (-b))^2 = (\sqrt{a^2-b^2})^2 \)
\( \implies (x + a)^2 + (y + b)^2 = a^2 - b^2 \)
Now, expand the squared terms:
\( (x^2 + 2ax + a^2) + (y^2 + 2by + b^2) = a^2 - b^2 \)
Rearrange the terms to get the general equation of the circle:
\( \implies x^2 + y^2 + 2ax + 2by + a^2 + b^2 - a^2 + b^2 = 0 \)
\( \implies x^2 + y^2 + 2ax + 2by + 2b^2 = 0 \)
This is the required equation of the circle. This form helps visualize the general shape of the circle easily.
In simple words: When you know the centre and the radius of a circle, you can put these numbers into a special formula. Then, you just open up the brackets and simplify to get the full equation of the circle.

🎯 Exam Tip: Remember the centre-radius form of the circle equation: \( (x-h)^2 + (y-k)^2 = r^2 \). Be careful with signs when substituting negative coordinates for the centre.

Question 3. Find the equation of the circle drawn on the line joining \( (-1, 2) \) and \( (3, -4) \) as diameter.
Answer:
We are given two points \( (x_1, y_1) = (-1, 2) \) and \( (x_2, y_2) = (3, -4) \) which are the endpoints of the diameter of a circle.
The equation of a circle when the endpoints of its diameter are known is given by the diameter form:
\( (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \)
Substitute the given coordinates into this formula:
\( (x - (-1))(x - 3) + (y - 2)(y - (-4)) = 0 \)
\( \implies (x + 1)(x - 3) + (y - 2)(y + 4) = 0 \)
Now, expand the terms:
\( (x^2 - 3x + x - 3) + (y^2 + 4y - 2y - 8) = 0 \)
\( \implies x^2 - 2x - 3 + y^2 + 2y - 8 = 0 \)
Combine the constant terms and rearrange:
\( \implies x^2 + y^2 - 2x + 2y - 11 = 0 \)
This is the required equation of the circle. The diameter form is very useful for this specific type of problem.
In simple words: If you have the two ends of a circle's diameter, you can use a special formula that puts these points into it. After multiplying everything out and simplifying, you get the equation of the circle.

🎯 Exam Tip: When given the diameter endpoints, use the specific diameter form \( (x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0 \). It's more direct than finding the centre and radius separately.

Question 4. Find the equation of the circle passing through the points \( (4, 1) \) and \( (6, 5) \) and whose centre lies on the line \( 4x + y = 16 \).
Answer:
Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \) ...(1)
The centre of this circle is \( (-g, -f) \).
We are given that the circle passes through the point \( (4, 1) \). Substitute \( x=4 \) and \( y=1 \) into equation (1):
\( (4)^2 + (1)^2 + 2g(4) + 2f(1) + c = 0 \)
\( 16 + 1 + 8g + 2f + c = 0 \)
\( \implies 8g + 2f + c + 17 = 0 \) ...(2)
The circle also passes through the point \( (6, 5) \). Substitute \( x=6 \) and \( y=5 \) into equation (1):
\( (6)^2 + (5)^2 + 2g(6) + 2f(5) + c = 0 \)
\( 36 + 25 + 12g + 10f + c = 0 \)
\( \implies 12g + 10f + c + 61 = 0 \) ...(3)
The centre \( (-g, -f) \) of the circle lies on the line \( 4x + y = 16 \). Substitute \( -g \) for \( x \) and \( -f \) for \( y \):
\( 4(-g) + (-f) = 16 \)
\( \implies -4g - f = 16 \)
\( \implies 4g + f + 16 = 0 \) ...(4)
Now we have a system of three equations (2), (3), and (4) with three unknowns (g, f, c). We need to solve for g, f, and c.
Subtract equation (2) from equation (3):
\( (12g + 10f + c + 61) - (8g + 2f + c + 17) = 0 \)
\( 4g + 8f + 44 = 0 \)
Divide by 4:
\( \implies g + 2f + 11 = 0 \) ...(5)
Now we have two equations, (4) and (5), involving g and f:
(4) \( 4g + f + 16 = 0 \)
(5) \( g + 2f + 11 = 0 \)
From (4), \( f = -4g - 16 \). Substitute this into (5):
\( g + 2(-4g - 16) + 11 = 0 \)
\( g - 8g - 32 + 11 = 0 \)
\( -7g - 21 = 0 \)
\( -7g = 21 \)
\( \implies g = -3 \)
Substitute \( g = -3 \) back into \( f = -4g - 16 \):
\( f = -4(-3) - 16 \)
\( f = 12 - 16 \)
\( \implies f = -4 \)
Finally, substitute the values of \( g = -3 \) and \( f = -4 \) into equation (2) to find c:
\( 8(-3) + 2(-4) + c + 17 = 0 \)
\( -24 - 8 + c + 17 = 0 \)
\( -32 + c + 17 = 0 \)
\( c - 15 = 0 \)
\( \implies c = 15 \)
Now, substitute the values of \( g = -3 \), \( f = -4 \), and \( c = 15 \) back into the general equation of the circle (1):
\( x^2 + y^2 + 2(-3)x + 2(-4)y + 15 = 0 \)
\( \implies x^2 + y^2 - 6x - 8y + 15 = 0 \)
This is the required equation of the circle. Finding the values of g, f, and c is key here.
In simple words: First, write down the general equation of a circle. Then, use the given points and the line where the centre lies to form three equations. Solve these equations to find the special numbers (g, f, c) for the circle. Finally, put these numbers back into the general equation to get your answer.

🎯 Exam Tip: For problems involving three conditions (like passing through two points and centre on a line), the general equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is usually the most efficient approach. Solving the system of linear equations correctly is crucial.

Question 5. Find the equation of a circle of radius 5 whose centre lies on x-axis and passes through the point \( (2, 3) \).
Answer:
Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \) ...(1)
The centre of the circle is \( (-g, -f) \).
Since the centre lies on the x-axis, its y-coordinate must be 0.
Therefore, \( -f = 0 \implies f = 0 \).
Substitute \( f=0 \) into equation (1):
\( x^2 + y^2 + 2gx + c = 0 \) ...(2)
We are given that the radius of the circle is 5.
The formula for the radius is \( r = \sqrt{g^2 + f^2 - c} \).
Substitute \( r=5 \) and \( f=0 \):
\( 5 = \sqrt{g^2 + 0^2 - c} \)
\( 5 = \sqrt{g^2 - c} \)
Squaring both sides:
\( 25 = g^2 - c \) ...(3)
We are also given that the circle passes through the point \( (2, 3) \). Substitute \( x=2 \) and \( y=3 \) into equation (2):
\( (2)^2 + (3)^2 + 2g(2) + c = 0 \)
\( 4 + 9 + 4g + c = 0 \)
\( \implies 4g + c + 13 = 0 \) ...(4)
Now we have a system of two equations, (3) and (4), with two unknowns (g and c):
(3) \( g^2 - c = 25 \)
(4) \( 4g + c + 13 = 0 \)
From (4), \( c = -4g - 13 \). Substitute this expression for c into equation (3):
\( g^2 - (-4g - 13) = 25 \)
\( g^2 + 4g + 13 = 25 \)
\( g^2 + 4g - 12 = 0 \)
This is a quadratic equation for g. We can factor it:
\( (g + 6)(g - 2) = 0 \)
This gives two possible values for g:
\( g = 2 \) or \( g = -6 \)

Case 1: When \( g = 2 \)
Substitute \( g=2 \) into \( c = -4g - 13 \):
\( c = -4(2) - 13 = -8 - 13 = -21 \)
Now substitute \( g=2 \), \( f=0 \), and \( c=-21 \) into equation (1):
\( x^2 + y^2 + 2(2)x + 2(0)y + (-21) = 0 \)
\( \implies x^2 + y^2 + 4x - 21 = 0 \) ...(5)

Case 2: When \( g = -6 \)
Substitute \( g=-6 \) into \( c = -4g - 13 \):
\( c = -4(-6) - 13 = 24 - 13 = 11 \)
Now substitute \( g=-6 \), \( f=0 \), and \( c=11 \) into equation (1):
\( x^2 + y^2 + 2(-6)x + 2(0)y + 11 = 0 \)
\( \implies x^2 + y^2 - 12x + 11 = 0 \) ...(6)
Thus, equations (5) and (6) are the two possible equations of circles that satisfy the given conditions. This shows that sometimes there can be more than one solution.
In simple words: Start with the general circle equation. Since the centre is on the x-axis, the 'f' value is zero. Use the given radius and the point the circle passes through to set up equations. Solve these equations to find 'g' and 'c', which might give you two different pairs of values. Each pair will give you one possible equation for the circle.

🎯 Exam Tip: When the centre lies on an axis, one coordinate (h or k) becomes 0, simplifying the general equation. Be prepared for multiple possible solutions when solving quadratic equations for g or f.

Question 6. Find the equation of the circle concentric with the circle \( x^2 + y^2 - 8x + 6y - 5 = 0 \) and passing through the point \( (-2, -7) \).
Answer:
We are given the equation of a circle: \( x^2 + y^2 - 8x + 6y - 5 = 0 \) ...(1)
To find the centre of this circle, we compare it with the general equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
From (1), we have \( 2g = -8 \implies g = -4 \) and \( 2f = 6 \implies f = 3 \).
The centre of circle (1) is \( (-g, -f) = (-(-4), -3) = (4, -3) \).
A concentric circle shares the same centre. So, the centre of the required circle is also \( (4, -3) \).
The required circle passes through the point \( (-2, -7) \).
The radius of the required circle will be the distance between its centre \( (4, -3) \) and the point it passes through \( (-2, -7) \). We use the distance formula \( r = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).
Let \( (x_1, y_1) = (4, -3) \) and \( (x_2, y_2) = (-2, -7) \).
Radius \( r = \sqrt{(-2 - 4)^2 + (-7 - (-3))^2} \)
\( \implies r = \sqrt{(-6)^2 + (-7 + 3)^2} \)
\( \implies r = \sqrt{36 + (-4)^2} \)
\( \implies r = \sqrt{36 + 16} \)
\( \implies r = \sqrt{52} \)
Now we have the centre \( (h, k) = (4, -3) \) and the radius \( r = \sqrt{52} \) for the required circle.
Using the centre-radius form of a circle's equation \( (x-h)^2 + (y-k)^2 = r^2 \):
\( (x - 4)^2 + (y - (-3))^2 = (\sqrt{52})^2 \)
\( \implies (x - 4)^2 + (y + 3)^2 = 52 \)
Expand the terms:
\( (x^2 - 8x + 16) + (y^2 + 6y + 9) = 52 \)
\( \implies x^2 + y^2 - 8x + 6y + 16 + 9 - 52 = 0 \)
\( \implies x^2 + y^2 - 8x + 6y - 27 = 0 \)
This is the required equation of the circle. Understanding what "concentric" means is key here.
In simple words: A concentric circle means it has the same centre as another circle. First, find the centre of the given circle. Then, use this same centre and the point the new circle passes through to calculate its radius. Finally, use the centre and the new radius to write the equation of the required circle.

🎯 Exam Tip: The term "concentric" means having the same centre. To find the centre \( (-g, -f) \) from \( x^2 + y^2 + 2gx + 2fy + c = 0 \), remember to change the signs of \( g \) and \( f \).

Question 7. Find the equation of the circle through the points \( (0, 0), (2, 0) \) and \( (0, 4) \). Also find the coordinates of its centre and its radius.
Answer:
Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \) ...(1)
The circle passes through the point \( (0, 0) \). Substitute \( x=0 \) and \( y=0 \) into equation (1):
\( (0)^2 + (0)^2 + 2g(0) + 2f(0) + c = 0 \)
\( \implies c = 0 \)
Now, the circle passes through the point \( (2, 0) \). Substitute \( x=2 \), \( y=0 \), and \( c=0 \) into equation (1):
\( (2)^2 + (0)^2 + 2g(2) + 2f(0) + 0 = 0 \)
\( 4 + 0 + 4g + 0 + 0 = 0 \)
\( 4 + 4g = 0 \)
\( 4g = -4 \)
\( \implies g = -1 \)
The circle also passes through the point \( (0, 4) \). Substitute \( x=0 \), \( y=4 \), and \( c=0 \) into equation (1):
\( (0)^2 + (4)^2 + 2g(0) + 2f(4) + 0 = 0 \)
\( 0 + 16 + 0 + 8f + 0 = 0 \)
\( 16 + 8f = 0 \)
\( 8f = -16 \)
\( \implies f = -2 \)
Now we have the values: \( g = -1 \), \( f = -2 \), and \( c = 0 \).
Substitute these values back into the general equation of the circle (1):
\( x^2 + y^2 + 2(-1)x + 2(-2)y + 0 = 0 \)
\( \implies x^2 + y^2 - 2x - 4y = 0 \)
This is the required equation of the circle. We can now find its centre and radius.
The centre of the circle is \( (-g, -f) \).
\( \implies \) Centre \( = (-(-1), -(-2)) = (1, 2) \)
The radius of the circle is \( r = \sqrt{g^2 + f^2 - c} \).
\( \implies r = \sqrt{(-1)^2 + (-2)^2 - 0} \)
\( \implies r = \sqrt{1 + 4 - 0} \)
\( \implies r = \sqrt{5} \)
So, the equation of the circle is \( x^2 + y^2 - 2x - 4y = 0 \), its centre is \( (1, 2) \), and its radius is \( \sqrt{5} \). This process is common for circles passing through three points.
In simple words: To find a circle's equation from three points, start with the general form. Plug in each point one by one to find the values for g, f, and c. Once you have these, you can write the circle's equation and then easily find its centre and radius using the standard formulas.

🎯 Exam Tip: When a circle passes through the origin \( (0,0) \), the constant term \( c \) in the general equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is always 0. This simplifies the calculations significantly.

Question 8. Find the parametric representation of the circle \( x^2 + y^2 - 2x - 4y - 4 = 0 \).
Answer:
To find the parametric representation, we first need to convert the given equation into the standard form \( (x-h)^2 + (y-k)^2 = r^2 \).
Given equation: \( x^2 + y^2 - 2x - 4y - 4 = 0 \)
Group the x-terms and y-terms together:
\( (x^2 - 2x) + (y^2 - 4y) = 4 \)
Now, complete the square for both the x-terms and y-terms. To complete \( x^2 - 2x \), we add \( (-\frac{2}{2})^2 = (-1)^2 = 1 \). To complete \( y^2 - 4y \), we add \( (-\frac{4}{2})^2 = (-2)^2 = 4 \). Remember to add these values to both sides of the equation.
\( (x^2 - 2x + 1) + (y^2 - 4y + 4) = 4 + 1 + 4 \)
\( \implies (x - 1)^2 + (y - 2)^2 = 9 \)
This is the standard form of the circle equation. From this, we can identify the centre \( (h, k) = (1, 2) \) and the radius \( r = \sqrt{9} = 3 \).
The parametric equations for a circle with centre \( (h, k) \) and radius \( r \) are given by:
\( x = h + r \cos \theta \)
\( y = k + r \sin \theta \)
Substitute \( h=1 \), \( k=2 \), and \( r=3 \):
\( x = 1 + 3 \cos \theta \)
\( y = 2 + 3 \sin \theta \)
Here, \( \theta \) (theta) is the parameter. These equations represent every point on the circle as \( \theta \) varies. Also, we can write the equation as \( \left(\frac{x-1}{3}\right)^2 + \left(\frac{y-2}{3}\right)^2 = 1 \), which clearly shows the cosine and sine relationships.
In simple words: First, change the circle's equation into its standard form, which shows the centre and radius. Once you have these, use the simple parametric formulas: x equals the x-coordinate of the centre plus radius times cosine of theta, and y equals the y-coordinate of the centre plus radius times sine of theta.

🎯 Exam Tip: To find parametric equations, always convert the given circle equation to the standard form \( (x-h)^2 + (y-k)^2 = r^2 \) by completing the square. Then, substitute h, k, and r into \( x = h + r \cos \theta \) and \( y = k + r \sin \theta \).

Question 9. Find the length of the chord intercepted by the circle \( x^2 + y^2 = 25 \) on the line \( 2x - y + 5 = 0 \).
Answer:
The equation of the circle is \( x^2 + y^2 = 25 \).
Comparing this with \( x^2 + y^2 = r^2 \), we see that the centre of the circle is \( C(0, 0) \) and the radius is \( R = \sqrt{25} = 5 \).
The equation of the line (chord) is \( 2x - y + 5 = 0 \).
To find the length of the chord, we first need to find the perpendicular distance (d) from the centre of the circle \( (0, 0) \) to the line \( 2x - y + 5 = 0 \).
The formula for the perpendicular distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \).
Here, \( (x_1, y_1) = (0, 0) \), \( A=2 \), \( B=-1 \), \( C=5 \).
\( d = \frac{|2(0) - 1(0) + 5|}{\sqrt{2^2 + (-1)^2}} \)
\( \implies d = \frac{|5|}{\sqrt{4 + 1}} = \frac{5}{\sqrt{5}} \)
To rationalize, multiply numerator and denominator by \( \sqrt{5} \):
\( \implies d = \frac{5\sqrt{5}}{5} = \sqrt{5} \)
Let the chord be AB, and M be the midpoint of the chord. The line segment CM (where C is the centre) is perpendicular to the chord AB. In the right-angled triangle CMA, where CA is the radius, CM is the perpendicular distance, and AM is half the chord length.
By the Pythagorean theorem: \( R^2 = d^2 + AM^2 \)
\( AM^2 = R^2 - d^2 \)
\( AM^2 = (5)^2 - (\sqrt{5})^2 \)
\( AM^2 = 25 - 5 \)
\( AM^2 = 20 \)
\( AM = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \)
The length of the chord AB is twice the length of AM.
Length of chord \( AB = 2 \times AM = 2 \times 2\sqrt{5} = 4\sqrt{5} \) units.
This shows how geometry and algebra work together for such problems.
In simple words: First, find the circle's centre and radius. Then, calculate how far the centre is from the line (this is 'd'). Imagine a right-angled triangle formed by the radius, 'd', and half the chord. Use a formula called Pythagoras theorem to find half the chord length, and then double it to get the full chord length.

🎯 Exam Tip: Always draw a simple diagram to visualize the problem. Remember that the perpendicular from the centre to a chord bisects the chord. The Pythagorean theorem is key for finding chord length after calculating the perpendicular distance.

Question 10. Find the equations of the tangents to the circle \( x^2 + y^2 = 9 \), which are parallel to the line \( 3x + 4y = 0 \).
Answer:
The equation of the given circle is \( x^2 + y^2 = 9 \) ...(1)
From this, we know that the centre of the circle is \( C(0, 0) \) and the radius is \( R = \sqrt{9} = 3 \).
The given line is \( 3x + 4y = 0 \) ...(2)
Any line parallel to \( 3x + 4y = 0 \) will have the form \( 3x + 4y + k = 0 \) ...(3), where k is a constant.
For line (3) to be a tangent to the circle (1), the perpendicular distance from the centre of the circle \( (0, 0) \) to the line must be equal to the radius \( R \).
Using the perpendicular distance formula \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \):
Here, \( (x_1, y_1) = (0, 0) \), \( A=3 \), \( B=4 \), \( C=k \), and \( d=R=3 \).
\( 3 = \frac{|3(0) + 4(0) + k|}{\sqrt{3^2 + 4^2}} \)
\( \implies 3 = \frac{|k|}{\sqrt{9 + 16}} \)
\( \implies 3 = \frac{|k|}{\sqrt{25}} \)
\( \implies 3 = \frac{|k|}{5} \)
\( \implies |k| = 15 \)
This means \( k = 15 \) or \( k = -15 \).
Substitute these two values of k back into the equation of the parallel line (3) to get the equations of the tangents:
When \( k = 15 \): \( 3x + 4y + 15 = 0 \)
When \( k = -15 \): \( 3x + 4y - 15 = 0 \)
These are the two required equations of the tangents. A circle generally has two tangents parallel to a given line.
In simple words: First, find the centre and radius of the circle. Then, write down the general form of a line that is parallel to the given line (it will look similar but have a new constant 'k'). For this parallel line to be a tangent, its distance from the circle's centre must be equal to the radius. Use this to find the values of 'k', and then write the final tangent equations.

🎯 Exam Tip: Lines parallel to \( Ax + By + C = 0 \) have the form \( Ax + By + k = 0 \). The condition for tangency is that the perpendicular distance from the centre to the line equals the radius. Don't forget to consider both positive and negative values for k.

Question 11. Find the equation of the circle which touches the y-axis at a distance of +4 from the origin and cuts off an intercept 6 from the x-axis.
Answer:
Since the circle touches the y-axis at a distance of +4 from the origin, the point of tangency is \( (0, 4) \).
This means the y-coordinate of the centre of the circle must be 4. Let the centre be \( (h, 4) \).
Because the circle touches the y-axis, the x-coordinate of the centre \( h \) is equal to the radius \( r \) (or \( -r \) if it's in the second quadrant). So, \( r = |h| \).
The equation of the circle can be written as \( (x-h)^2 + (y-4)^2 = h^2 \).
The circle cuts off an intercept of 6 from the x-axis. This means when \( y=0 \), the difference between the x-coordinates of the intersection points is 6.
Substitute \( y=0 \) into the circle's equation:
\( (x-h)^2 + (0-4)^2 = h^2 \)
\( (x-h)^2 + 16 = h^2 \)
\( x^2 - 2hx + h^2 + 16 = h^2 \)
\( x^2 - 2hx + 16 = 0 \)
Let \( x_1 \) and \( x_2 \) be the roots of this quadratic equation, which are the x-coordinates where the circle intersects the x-axis. The length of the intercept is \( |x_2 - x_1| = 6 \).
We know that for a quadratic equation \( ax^2 + bx + c = 0 \), the difference between roots is \( \sqrt{b^2 - 4ac}/|a| \).
Here, \( a=1 \), \( b=-2h \), \( c=16 \).
\( |x_2 - x_1| = \frac{\sqrt{(-2h)^2 - 4(1)(16)}}{1} = \sqrt{4h^2 - 64} \)
We are given that the intercept is 6:
\( 6 = \sqrt{4h^2 - 64} \)
Square both sides:
\( 36 = 4h^2 - 64 \)
\( 100 = 4h^2 \)
\( h^2 = 25 \)
\( \implies h = \pm 5 \)
This gives two possible values for \( h \), meaning there are two such circles.

Case 1: When \( h = 5 \)
The centre is \( (5, 4) \) and the radius \( r = |5| = 5 \).
Equation of the circle: \( (x-5)^2 + (y-4)^2 = 5^2 \)
\( (x-5)^2 + (y-4)^2 = 25 \)
Expand this:
\( x^2 - 10x + 25 + y^2 - 8y + 16 = 25 \)
\( \implies x^2 + y^2 - 10x - 8y + 16 = 0 \)

Case 2: When \( h = -5 \)
The centre is \( (-5, 4) \) and the radius \( r = |-5| = 5 \).
Equation of the circle: \( (x-(-5))^2 + (y-4)^2 = 5^2 \)
\( (x+5)^2 + (y-4)^2 = 25 \)
Expand this:
\( x^2 + 10x + 25 + y^2 - 8y + 16 = 25 \)
\( \implies x^2 + y^2 + 10x - 8y + 16 = 0 \)
These are the two equations of the circles that satisfy the given conditions. The existence of two solutions is common for such geometric problems involving tangency and intercepts.
In simple words: First, figure out the centre's y-coordinate and radius based on where the circle touches the y-axis. Then, use the information about the x-axis intercept to set up an equation that helps you find the centre's x-coordinate. You might find two possible x-coordinates, leading to two different circles, so write the equation for each one.

🎯 Exam Tip: When a circle touches an axis, the radius is equal to the absolute value of the coordinate of the centre perpendicular to that axis. For intercepts, remember that the length of the intercept can be related to the radius and the distance of the centre from the axis.