Question

$C_1=x^2+y^2-4x-8y-5=0$ and $C_2=x^2+y^2-12x-2y+12=0$ are the equations of two given circles. The equation of the circle on the common chord $PQ$ as diameter is:

• A. $2x^2+2y^2-16x-10y-27=0$
• B. $x^2+y^2-16x-10y-27=0$
• C. $x^2+y^2+16x-10y-27=0$
• D. $2x^2+2y^2+16x+10y-27=0$

Answer 1

The correct option is A $2x^2+2y^2-16x-10y-27=0$

The common chord is diameter $\Rightarrow$ The center should lie on the common chord.

Equation of common chord is $C_1-C_2=0\Rightarrow 8x-6y-17=0$

The equation of the circle passing through the extremities of the common chord is
$x^2+y^2-4x-8y-5+\lambda (8x-6y-17)=0$ ... (1)

Its centre is $(2-4\lambda ,4+3\lambda )$, This should lie on $PQ$.

$\Rightarrow 8(2-4\lambda )-6(4+3\lambda )-17=0$

$\Rightarrow 16-32\lambda -24-18\lambda -17=0$

$\Rightarrow -50\lambda =25$

$\Rightarrow \lambda =-\frac{25}{50}=-\frac{1}{2}$

$\therefore$ The required equation of the circle from (1) is:

$\Rightarrow$ $x^2+y^2-4x-8y-5+\frac{-1}{2}(8x-6y-17)=0$

$\Rightarrow$$2x^2+2y^2-16x-10y-27=0$