Question

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

Answer 1

We know that the equation of circle passing through intersection of given circles is $\Rightarrow (x^2+y^2-2x-6y+6)+k(x^2+y^2+2x-6y)=0$

$\Rightarrow (1+k)x^2+(1+k)y^2+(-2+2k)x+(-6-6k)y+6=0$

$\Rightarrow (1+k)x^2+(1+k)y^2+2(k-1)x+2(-3-3k)y+6=0$ ....$\left(1\right)$

$\Rightarrow x^2+y^2+2\left(\frac{k-1}{k+1}\right)x+2\left(\frac{-3-3k}{k+1}\right)y+\frac{6}{k+1}=0$

$\Rightarrow x^2+y^2+2\left(\frac{k-1}{k+1}\right)x+2(-3)y+\frac{6}{k+1}=0$

Given that this circle intersects $x^2+y^2+4x+6y+4=0$ orthogonally.

The condition or orthogonal intersection of 2 circles is $2(g{g}^{\prime }+f{f}^{\prime })=c+c^{\prime }$

$\Rightarrow 2(\frac{k-1}{k+1}\times 2+(-3\left)\right(3\left)\right)=4+\frac{6}{k+1}$

$\Rightarrow \frac{2(k-1)}{k+1}-9=2+\frac{3}{k+1}$

$\Rightarrow \frac{2(k-1)}{k+1}-11=\frac{3}{k+1}$

$\Rightarrow 2(k-1)-11(k+1)=3$

On solving we get, $k=\frac{16}{9}$

Substituting this value of $k$ in $\left(1\right)$ we get,

$\Rightarrow 25x^2+25y^2+14x-150y+54=0$