The equation of the circle which passes through the centre of the circle $x^2+y^2+8 x+10 y-7=0$ and…
The equation of the circle which passes through the centre of the circle $x^2+y^2+8 x+10 y-7=0$ and concentric which the circle $2 x^2+2 y^2-8 x-12 y-9=0$ is
$x^2+y^2-4 x-6 y+77=0$
$x^2+y^2-4 x-6 y-89=0$
$x^2+y^2-4 x-6 y+97=0$
$x^2+y^2-4 x-6 y-87=0$
Solution
Required circle is concentric with
$\begin{aligned}
& 2 x^2+2 y^2-8 x-12 y-9=0 \\
& \Rightarrow x^2+y^2-4 x-6 y-\frac{9}{2}=0
\end{aligned}$
$\therefore \quad$ Centre is $(2,3)$
Also, it passes through centre of
$x^2+y^2+8 x+10 y-7=0$
$\therefore \quad$ Centre is $(-4,-5)$
$\therefore \quad$ Radius $=\sqrt{(-4-2)^2+(-5-3)^2}=10$
$\therefore \quad$ Equation of required circle is
$\begin{aligned} & (x-2)^2+(y-3)^2=10^2 \\ & x^2+y^2-4 x-6 y-87=0\end{aligned}$