Question

The equation of the circle passing through the origin and the points of intersection of the circles $x^2+y^2-4x-6y-3=0$, $x^2+y^2+4x-2y-4=0$

• A. $x^2+y^2+28x+18y=0$
• B. $x^2+y^2-18x-28y=0$
• C. $x^2+y^2-28x+18y=0$
• D. $x^2+y^2-28x-18y=0$

Answer 1

The correct option is D $x^2+y^2-28x-18y=0$

By family of circles,
equation of circle passing through point of intersection
of $s_1:x^2+y^2-4x-6y-3=0$ and $s_2:x^2+y^2+4x-2y-4=0$ is
$\therefore s_1+\lambda s_2=0$
$\Rightarrow (1+\lambda )x^2+(1+\lambda )y^2+4(\lambda -1)x-2(3+\lambda )y-3-4\lambda =0$
Given this circles pass through origin $(0,0)$
$\therefore -3-4\lambda =0$
$\lambda =\frac{-3}{4}$
$\therefore$ Equation of circle is
$\frac{x^2}{4}+\frac{y^2}{4}-\frac{28x}{4}-\frac{18y}{4}=0$
$\Rightarrow x^2+y^2-28x-18y=0$