Discussion Forum

The equation  of the circle passing through the foci of the ellipse 

$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$  and having centre at (0,3) is 

Answer:

Explanation:

Given  equation of ellipse is  $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$  

 Here, a=4,b=3,   $e=\sqrt{1-\frac{9}{16}}\Rightarrow\frac{\sqrt{7}}{4}$

 $\therefore$  Foci is (±,0)= $\left(\pm4\times \frac{\sqrt{7}}{4},0\right)$

$= \left(\pm \sqrt{7},0\right)$

$\therefore$   Radius of the circle,

           $r= \sqrt{(ae)^{2}+b^{2}}=\sqrt{7+9}=\sqrt{16}=4$

 Now, equation of circle is 

                      $(x-0)^{2}+(y-3)^{2}=16$

  $\therefore$   $x^{2}+y^{2}-6y-7=0$