Question

The equation of the circle passing through (1, 1) and the points of intersection of $x^2+y^2+13x-3y=0$ and $2x^2+2y^2+4x-7y-25=0$ is

• A. $2x^2+2y^2+25x-13y-30=0$
• B. $4x^2+4y^2+30x-13y-25=0$
• C. $x^2+y^2+15x-13y-25=0$
• D. $4x^2+4y^2+25x-13y-30=0$

Answer 1

The correct option is B $4x^2+4y^2+30x-13y-25=0$

Circle through points of intersection of given two circles is
$(x^2+y^2+13x-3y)+\lambda (2x^2+2y^2+4x-7y-25)=0\Rightarrow (1+2\lambda )x^2+(1+2\lambda )y^2+(13+4\lambda )x+(-3-7\lambda )y-25\lambda =0$
As it passes through (1, 1),
$\therefore 1+2\lambda +1+2\lambda +13+4\lambda -3-7\lambda -25\lambda =0\Rightarrow -24\lambda +12=0\Rightarrow \lambda =\frac{1}{2}\therefore Requiredcircleis2x^2+2y^2+15x-\frac{13y}{2}-\frac{25}{2}=0or4x^2+4y^2+30x-13y-25=0$