If the curves $\frac{{\mathrm{x}}^2}{4}+{\mathrm{y}}^2=1\text{ and }\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+{\mathrm{y}}^2=1$ for a suitable value of a cut on four concyclic points, the equation of the circle passing through these four points is

The equation of conic through the point of intersection of given two ellipses is $$\begin{gathered} \left(\frac{{\mathrm{x}}^2}{4}+{\mathrm{y}}^2-1\right)+\mathrm{\lambda }\left(\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+{\mathrm{y}}^2-1\right)=0 \\ \text{ ⇒ }{\mathrm{x}}^2\left(\frac{1}{4}+\frac{\mathrm{\lambda }}{{\mathrm{a}}^2}\right)+{\mathrm{y}}^2(1+\mathrm{\lambda })=1+\mathrm{\lambda } \\ \text{ ⇒}{\mathrm{x}}^2\left(\frac{{\mathrm{a}}^2+4\mathrm{\lambda }}{4{\mathrm{a}}^2(1+\mathrm{\lambda })}\right)+{\mathrm{y}}^2=1 \end{gathered}$$
$$\begin{gathered} \text{ This equation is a circle if }\frac{{\mathrm{a}}^2+4\mathrm{\lambda }}{4{\mathrm{a}}^2(1+\mathrm{\lambda })}=1 \\ \text{ Therefore, the circle is }{\mathrm{x}}^2+{\mathrm{y}}^2=1\text{. } \end{gathered}$$