Question

Find the equation of the circle passing through the points of intersection of the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+4y-12=0$ and intersection the circle $x^2+y^2-2x-4=0$ orthogonally.

Answer 1

Let, the equation of the circle through the intersection of given circle is,

$x^2+y^2-4x-6y+12=0⟶\left(1\right)$

Here, the equation of radical axis for the circles

$x^2+y^2-4x-6y+12=0$

and $x^2+y^2+6x-+4y-12=0$ is,

$-10x-10y=0$

So, equation $\left(1\right)$ can be written as,

$x^2+y^2-4x-6y-12+\lambda (-10x-10y)=0\Rightarrow x^2+y^2-x(10\lambda +4)-y(10\lambda +6)-12=0$

It cuts the circle $v$ orthogonally.

$\therefore 2g{g}_1+2f{f}_1=c+c_1\Rightarrow (10\lambda +4)\left(1\right)+(10\lambda +6)\left(0\right)=-12-4\Rightarrow \lambda =-2$

Hence, the required equation of the circle is,

$x^2+y^2-4x-6y-12-2(-10x-10y)=0\Rightarrow x^2+y^2+16x+14y-12=0.$