Question

The equation of the circle having one of its diameter as end points of foci of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$, is

• A. $x^2+y^2=a^2+b^2$
• B. $x^2+y^2=a^2$
• C. $x^2+y^2=2a^2$
• D. $x^2+y^2=a^2-b^2$

Answer 1

The correct option is D $x^2+y^2=a^2-b^2$

Coordinates of focii of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are $(\pm ae,0)$
So centre of required circle is $(0,0)$ with radius $ae$ units.

Hence, the equation of the circle is $x^2+y^2=a^2{e}^2$
$\Rightarrow x^2+y^2=a^2(1-\frac{b^2}{a^2})$
$\Rightarrow x^2+y^2=a^2-b^2$