​The equation of the circle passing through the points of intersection of  the ellipse's x^2/a^2+y^2/b^2=1  and x^2/b^2+y^2/a^2=1 is-

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Please find below the solution to the asked query :
$$\begin{gathered} \mathrm{The} \mathrm{required} \mathrm{circle} \mathrm{will} \mathrm{be}, \\ \left(\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}-1\right)+\mathrm{\lambda }\left(\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}-1\right)= 0 \\ {\mathrm{x}}^2\left(\frac{1}{{\mathrm{a}}^2}+\frac{\mathrm{\lambda }}{{\mathrm{b}}^2}\right)+{\mathrm{y}}^2\left(\frac{1}{{\mathrm{b}}^2}+\frac{\mathrm{\lambda }}{{\mathrm{a}}^2}\right)-1-\mathrm{\lambda }=0 \\ \mathrm{Now} \mathrm{for} \mathrm{circle}, \mathrm{coefficient} \mathrm{of} {\mathrm{x}}^2 \mathrm{and} {\mathrm{y}}^2 \mathrm{must} \mathrm{be} \mathrm{same} \mathrm{so} \\ \left(\frac{1}{{\mathrm{a}}^2}+\frac{\mathrm{\lambda }}{{\mathrm{b}}^2}\right)= \left(\frac{1}{{\mathrm{b}}^2}+\frac{\mathrm{\lambda }}{{\mathrm{a}}^2}\right) \\ \mathrm{\lambda }\left(\frac{1}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}\right)= \left(\frac{1}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}\right) \\ \mathrm{\lambda }= 1, \mathrm{putting} \mathrm{this} \mathrm{we} \mathrm{get} \\ {\mathrm{x}}^2\left(\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}\right)+{\mathrm{y}}^2\left(\frac{1}{{\mathrm{b}}^2}+\frac{1}{{\mathrm{a}}^2}\right)-2=0 \\ {\mathrm{x}}^2+{\mathrm{y}}^2= \frac{2}{\left(\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}\right)} \end{gathered}$$

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