The equation of circle passing through $(0,0)$ and cutting orthogonally the circles $x^2+y^2+6 x-15=0$ and…
The equation of circle passing through $(0,0)$ and cutting orthogonally the circles $x^2+y^2+6 x-15=0$ and $x^2+y^2-8 y-10=0$ is
$2\left(x^2+y^2\right)-10 x+5 y=0$
$2\left(x^2+y^2\right)+10 x-5 y=0$
$2\left(x^2-y^2\right)+10 x+5 y=0$
$2\left(x^2-y^2\right)-10 x-5 y=0$
Solution
Let the equation of the circle passing through $(0,0)$ is
$x^2+y^2+2 g x+2 f y=0$ ...(i)
and the given equations are
$x^2+y^2+6 x-15=0$ ...(ii)
$x^2+y^2-8 y-10=0$ ...(iii)
Eqs. (i) and (ii) are orthogonal.
$\begin{aligned} \therefore \quad 2 g \times 3+2 f \times 0 & =0+(-15) \\ & {\left[\therefore 2 g_1 g_2+2 f_1 f_2=c_1+c_2\right] }\end{aligned}$
$\Rightarrow \quad g=-\frac{5}{2}$
$\therefore$ Again Eqs. (i) and (iii) are orthogonal
$2 g \times 0+2 f \times(-4)=0+(-10)$
$\Rightarrow \quad f=5 / 4$
$\therefore$ The equation of the circle will be
$x^2+y^2+2\left(-\frac{5}{2}\right) x+2\left(\frac{5}{4}\right) y=0$
$\Rightarrow \quad 2\left(x^2+y^2\right)-10 x+5 y=0$