The equation of a circle, which passes through the centre of the circle $x^2+y^2+8 x+10 y-7=0$ and is…
The equation of a circle, which passes through the centre of the circle $x^2+y^2+8 x+10 y-7=0$ and is concentric with the circle. $2 x^2+2 y^2-8 x-12 y-9=0$, is
$x^2+y^2-4 x+6 y-87=0$
$x^2+y^2+4 x+6 y-87=0$
$x^2+y^2+4 x+6 y+87=0$
$x^2+y^2-4 x-6 y-87=0$
Solution
The required circle is concentric with the circle
$2 x^2+2 y^2-8 x-12 y-9=0$
i.e, centre is at $(2,3)$ and passes through the centre of the circle
$x^2+y^2+8 x+10 y-7=0$
i.e., through $(-4,-5)$
Hence, the required equation is
$\begin{aligned}
& (x-2)^2+(y-3)^2=(2+4)^2+(3+5)^2 \\
& \Rightarrow x^2+y^2-4 x-6 y-87=0
\end{aligned}$