The equation of the circle which passes through the point $(3,2)$ bisects the circumference of the circle…
The equation of the circle which passes through the point $(3,2)$ bisects the circumference of the circle $x^2+y^2=15$ and cuts the circle $x^2+y^2+4 x+6 y+3=0$ orthogonally is
$x^2+y^2+6 x+8 y-43=0$
$x^2+y^2+6 x-8 y-15=0$
$x^2+y^2-6 x+8 y-11=0$
$x^2+y^2-6 x-8 y+21=0$
Solution
Let the equation of required circle is
$
x^2+y^2+2 g x+2 f y+c=0
$
Since, circle (i) passes through point $(3,2)$, so
$
\begin{aligned}
& & 9+4+6 g+4 f+c & =0 \\
\Rightarrow & & 6 g+4 f+c+13 & =0
\end{aligned}
$
Since, circle (i) bisects the circumference of the circle $x^2+y^2=15$, so the common chord passes through the centre of the circle $x^2+y^2=15$.
So,
$
\begin{aligned}
c+15 & =0 \\
c & =-15
\end{aligned}
$
Since, circle Eq. (i) cuts the circle
$
x^2+y^2+4 x+6 y+3=0
$
Orthogonally, so
$
4 g+6 f=c+3
$
From Eqs. (ii), (iii) and (iv)
$
g=3, f=-4, c=-15
$
So, required equation of circle is
$
x^2+y^2+6 x-8 y-15=0 .
$