Question

Find the equation of the circle which cuts the following circles orthogonally.
(i) $x^2+y^2+4x-7=0$
$2x^2+2y^2+3x+5y-9=0$
$x^2+y^2+y=0$.

Answer 1

Let the circle required be

$x^2+y^2+2gx+2fy+c=0$

We know two circles intersect orthogonal

if:

$2g_1{g}_2+2f_1{f}_2=c_1+c_2$

Hence, here :

$2(-2)\left(g\right)+2\left(0\right)=c-7$

$\Rightarrow -4g=c-7$

$\Rightarrow c+4g=7\to \left(1\right)$

Then, $2\left(\frac{-3}{4}\right)\left(g\right)+2\left(\frac{-5}{4}\right)\left(f\right)=c-9$

$\Rightarrow -\frac{3}{2}g-\frac{5}{2}f=c-9$

$\Rightarrow -3g-5f=2c-18$

$\Rightarrow 2c+3g+5f=18\to \left(2\right)$

Also, $2\left(0\right)\left(g\right)+2\left(\frac{-1}{2}\right)\left(f\right)=c+0$

$\Rightarrow -f=c\to \left(3\right)$

Putting $\left(3\right)$ in $\left(1\right)$

$\Rightarrow 4g-f=7\to \left(4\right)$

Putting $\left(3\right)$ in $\left(2\right)$

$\Rightarrow -2f+3g+5f=18$

$\Rightarrow 3g+3f=18$

$\Rightarrow g+f=6\to \left(5\right)$

Solving $\left(4\right)$ and $\left(5\right)$

$\Rightarrow 4g-f=7+(g+f=6)\Rightarrow 5g=13$

$\Rightarrow g=\frac{13}{5}$

$f=6-g=6-\frac{13}{5}=\frac{17}{5}$

$c=-f=-\frac{17}{5}$

$\therefore$ Required circle

$x^2+y^2+\frac{26}{5}x+\frac{34}{5}y-\frac{17}{5}=0$