Question

The circle $x^2+y^2=4$ cuts the circle $x^2+y^2+2x+3y-5=0$ in $A$ & $B$. Then the equation of the circle on $AB$ as a diameter is:

• A. $13\left(x^2+y^2\right)-4x-6y-50=0$
• B. $9\left(x^2+y^2\right)+8x-4y+25=0$
• C. $x^2+y^2-5x+2y+72=0$
• D. None of these

Answer 1

Answer: (A)

Common chord of given circle $2x+3y-1=0$ family of circle passing through point of intersection of given circle
$\left(x^2+y^2+2x+3y-5\right)+\lambda \left(x^2+y^2-4\right)=0$
$(\lambda +1)x^2+(\lambda +1)y^2+2x+3y-(4\lambda +5)=0$
$x^2+y^2+\frac{2x}{\lambda +1}+\frac{3}{\lambda +1}y-\frac{(4\lambda +5)}{\lambda +1}=0$

centre $\left(-\frac{1}{\lambda +1},\frac{-3}{2(\lambda +1)}\right)$
This centre lies on
$2\left(-\frac{1}{\lambda +1}\right)+3\left(\frac{-3}{2(\lambda +1)}\right)-1=0$
$-4-9-2\lambda -2=0$
$\Rightarrow 2\lambda =-15\Rightarrow \lambda =-15/2$
$\left(-\frac{15}{2}+1\right)x^2+\left(-\frac{15}{2}+1\right)y^2+2x+3y-\left(-4\times \frac{15}{2}+5\right)=0$
$\Rightarrow -\frac{13x^2}{2}-\frac{13y^2}{2}+2x+3y+25=0$
$\Rightarrow 13\left(x^2+y^2\right)-4x-6y-50=0$