Question

The equation of the circle drawn with the two foci of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as the end-points of a diameter is
(a) x² + y² = a² + b²
(b) x² + y² = a²
(c) x² + y² = 2a²
(d) x² + y² = a² − b²

Answer 1

$$\begin{gathered} \left(\mathrm{d}\right){\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2-{\mathrm{b}}^2 \\ \mathrm{We}\mathrm{have}r=ae \\ \text{Let the equation of the circle be}{x}^2+y^2=r^2. \\ \mathrm{Now},x^2+y^2=a^2{e}^2\left(\because r=ae\right) \\ \Rightarrow x^2+y^2=a^2\left(1-\frac{b^2}{a^2}\right)\left(\because e=\sqrt{1-\frac{b^2}{a^2}}\right) \\ \Rightarrow x^2+y^2=a^2-b^2 \\ \therefore \text{The required equation of the circle is}{x}^2+y^2=a^2-b^2. \end{gathered}$$