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

पर्याय

• x² + y² = a² + b²

• x² + y² = a²

• x² + y² = 2a²

• x² + y² = a² − b²

Answer 1

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