Problem source

Show that the equation of any circle passing through the points of intersection of the ellipse  $(x+2)^2 +2y^2 =18$ and the ellipse $9(x-1)^2  +16y^2 =  25$ can be written in the form
\[
x^2-2ax +y^2 =5-4a\;.
\]

Solution source

\begin{align*}
&& (x+2)^2 +2y^2 &=18 \\
&& 9(x-1)^2  +16y^2 &=  25 \\
\Rightarrow && 2y^2 &= 18 - (x+2)^2 \\
&& 16y^2 &= 25 - 9(x-1)^2 \\
\Rightarrow && 25-9(x-1)^2 &= 8 \cdot 18 - 8(x+2)^2 \\
\Rightarrow && 25 -9+18x-9x^2 &= 144 -32- 32x  +8x^2 \\
\Rightarrow && 0 &= 96 - 50x+x^2 \\
&&&= (x-48)(x-2) \\
\Rightarrow && x &= 2,48 \\
\Rightarrow && 2y^2 &= 2, 18-50^2 \\
\Rightarrow && (x,y) &= (2,\pm1)
\end{align*}

Therefore any circle must have it's centre on there perpendicular bisector of $(2, \pm 1)$, ie on the $x$-axis.

Therefore it will have equation $(x-a)^2+y^2 = r^2$ and also contain the point $(2,1)$, therefore:

\begin{align*}
r^2 &= (2-a)^2 + 1^2 \\
&= 4 -2a+a^2 + 1 \\
&= 5-2a+a^2
\end{align*}

and the equation is:

\begin{align*}
&& (x-a)^2 + y^2 &= 5-4a+a^2 \\
\Rightarrow && x^2-2ax+a^2 +y^2 &= 5-4a+a^2 \\
\Rightarrow && x^2-2ax+y^2 &= 5-4a
\end{align*}

as required.