Explanation

(d)
Given equation of ellipse are
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$
[image:11_image8.emf]
The point of the intersection of these ellipse are
$\left( \pm \frac{ab}{\sqrt{a^{2} + b^{2}}},\ \pm \frac{ab}{\sqrt{a^{2} + b^{2}}} \right)ie,\$
$P\left( \frac{ab}{\sqrt{a^{2} + b^{2}}},\ \frac{ab}{\sqrt{a^{2} + b^{2}}} \right)$
$\therefore$ The distance between $OP = r$
$= \sqrt{\left( \frac{ab}{\sqrt{a^{2} + b^{2}}} - 0 \right)^{2} + \left( \frac{ab}{\sqrt{a^{2} + b^{2}}} - 0 \right)^{2}}$
$= \frac{ab}{\sqrt{a^{2} + b^{2}}}\sqrt{2}$
$\therefore$ Equation of circle is
$x^{2} + y^{2} = r^{2}$
$\Rightarrow x^{2} + y^{2} = \frac{{2a}^{2}b^{2}}{a^{2} + b^{2}}$