Question

The equation of the circle passing through $(0,0)$ and cutting the circles
$x^2+y^2+6x-15=0,x^2+y^2-8y+10=0$ orthogonally is

• A. $(x-\frac{5}{2}{)}^2+(y-\frac{5}{4}{)}^2=\frac{125}{16}$
• B. $x^2+y^2-5x-5y=0$
• C. $2(x^2+y^2)-10x-5y=0$
• D. $x^2+y^2-5x+5y=0$

Answer 1

Answer: (A), (C)

The correct options are
A $(x-\frac{5}{2}{)}^2+(y-\frac{5}{4}{)}^2=\frac{125}{16}$
C $2(x^2+y^2)-10x-5y=0$

Let $S:x^2+y^2+2gx+2fy=0$ is a circle passing through $(0,0)$
then $s_1:x^2+y^2+6x-15=0$ and $s_2:x^2+y^2-8y+10=0$ are orthogonally to $s$.
$\Rightarrow 2g_1{g}_2+2f_1{f}_2=c_1+c_2$ [orthogonal circles property]
then, $2g\times \left(3\right)+2f\left(0\right)=0-15$
$g=\frac{-5}{2}$
and $2g\left(0\right)+2f(-4)=0+10$
$f=\frac{-5}{4}$
So, $e{q}^n$ of circle is $x^2+y^2-5x-\frac{5}{2}y=0$

option $A$ and $C$ are similiar