Question

If the circle $x^2+y^2+2gx+2fy+c=0$ bisects the circumference of the circle $x^2+y^2+2g^{\prime }x+2f^{\prime }y+c^{\prime }$, then the length ofthe common chord of these two circles is

• A. $2\sqrt{g^2+f^2-c}$
• B. $2\sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime }}$
• C. $2\sqrt{g^2+f^2+c}$
• D. $2\sqrt{g^{\prime 2}+f^{\prime 2}+c^{\prime }}$

Answer 1

The correct option is B $2\sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime }}$

The given circles are

$S_1:x^2+y^2+2gx+2fy+c=0\cdots \left(1\right)$

$S_2:x^2+y^2+2g^{\prime }x+2f^{\prime }y+c^{\prime }=0\cdots \left(2\right)$

As $\left(1\right)$ bisects the circumference of $\left(2\right)$ So the the common chord must be the diameter of circle $\left(2\right)$

So diameter of circle $\left(2\right)$ is $2\sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime }}$

Hence the length of common chord is $2\sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime }}$

The answer is $\text{Opt : B}$