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. $x^2+y^2=a^2+b^2$
  
• 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 $x^2+y^2=a^2-b^2$

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