Question

A circle passing through the point (1,0) makes an intercept of length 4 units on X-axis and an intercept of length \(2\sqrt{11}\) units on Y-axis. If the centre of the circle lies in the fourth quadrant, then the radius of the circle is

• A. \( 4\sqrt{5} \)
• B. 3
• C. \( 2\sqrt{5} \)
• D. 5

Hint:

Circle \(x^2+y^2+2gx+2fy+c=0\): Centre \( (-g,-f) \), Radius \( \sqrt{g^2+f^2-c} \). X-intercept \( = 2\sqrt{g^2-c} \). Y-intercept \( = 2\sqrt{f^2-c} \). Use given conditions to form system of equations for \(g,f,c\). Quadrant info constrains signs of \(g,f\).

Answer 1

The Correct Option is C

Let the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
Centre \( (-g, -f) \), Radius \( R = \sqrt{g^2 + f^2 - c} \).
Passes through \( (1, 0) \): \( 1 + 0 + 2g(1) + 0 + c = 0 \implies 1 + 2g + c = 0 \quad \cdots (1) \).
X-intercept length: \( 2\sqrt{g^2 - c} = 4 \implies g^2 - c = 4 \quad \cdots (2) \).
Y-intercept length: \( 2\sqrt{f^2 - c} = 2\sqrt{11} \implies f^2 - c = 11 \quad \cdots (3) \).
Centre \( (-g, -f) \) in the 4th quadrant: \( -g > 0 \implies g < 0 \); and \( -f < 0 \implies f > 0 \).
From (2), \( c = g^2 - 4 \).
Substitute into (1): \[ 1 + 2g + (g^2 - 4) = 0 \implies g^2 + 2g - 3 = 0 \implies (g + 3)(g - 1) = 0 \] Since \( g < 0 \), we take \( g = -3 \).
Then \( c = (-3)^2 - 4 = 9 - 4 = 5 \).
Substitute \( c = 5 \) into (3): \[ f^2 - 5 = 11 \implies f^2 = 16 \implies f = \pm 4 \] Since \( f > 0 \), we take \( f = 4 \).
So, \( g = -3, f = 4, c = 5 \).
Centre is \( (3, -4) \), which is in the 4th quadrant.
Radius \( R = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + 4^2 - 5} = \sqrt{9 + 16 - 5} = \sqrt{25 - 5} = \sqrt{20} \).
\[ R = \sqrt{4 \times 5} = 2\sqrt{5} \] This matches option (3).