The points of intersection of the two ellipse ${\mathrm{x}}^2+2{\mathrm{y}}^2-6\mathrm{x}-12y+23=0,4{\mathrm{x}}^2+2{\mathrm{y}}^2-20\mathrm{x}-12\mathrm{y}+35=0$

Given ellipse are

 $$\begin{gathered} 4{\mathrm{x}}^2+2{\mathrm{y}}^2-20\mathrm{x}-12\mathrm{y}+35=0\to \left(1\right) \\ and {\mathrm{x}}^2+2{\mathrm{y}}^2-6\mathrm{x}-12\mathrm{y}+23=0\to \left(2\right) \end{gathered}$$

Equation of the curve through the intersection of (1) and (2) is $4{\mathrm{x}}^2+2{\mathrm{y}}^2-20\mathrm{x}-12\mathrm{y}+35+\mathrm{\lambda }\left({\mathrm{x}}^2+2{\mathrm{y}}^2-6\mathrm{x}-12\mathrm{y}+23\right)=0$
If it represents a circle then $4+\mathrm{\lambda }=2+2\mathrm{\lambda }\Rightarrow \mathrm{\lambda }=2$
 $$\begin{gathered} then circle equation becomes 6x^2+6y^2-32x-36y+81=0 \\ \Rightarrow x^2+y^2-\left(\frac{16}{3}\right)x-6y+\frac{81}{6}=0 \\ centre of ellipse=(\frac{8}{3},3) \\ and radius=\sqrt{\frac{64}{9}+9-\frac{81}{6}} \\ =\frac{1}{3}\sqrt{\frac{47}{2}} \end{gathered}$$