Question

the radius of the circle passing through the foci of the ellipse

x²/16 + y²/9 = 1 and having its centre at (0,3)?

Answer 1

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

$\text{Comparing this with }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ we get}$

$a^2=16\;\;\&\;\;b^2=9$

$ae=\sqrt{a^2-b^2}$

$ae=\sqrt{16-9}$

$ae=\sqrt{7}$

$\text{Foci are ( \pm\sqrt{7} , 0)}$

$\text{since the circle passes through ( \sqrt{7} , 0),}$

$\text{Radius of the circle=}\text{Distance between centre and one of the foci}$

$\text{Radius=}\sqrt{(\sqrt7-0)^2+(0-3)^2}$

$\text{Radius=}\sqrt{7+9}$

$\text{Radius=}4\text{ units}$

$\therefore\textbf{Radius of the required circle is 4 units}$