The equation of the circle passing through the foci of the ellipse x^216+y^29=1, and havin…

Question

The equation of the circle passing through the foci of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$, and having centre at $(0,3)$, is:

Accepted Answer

From the given equation of ellipse, we have

$$a = 4,b = 3,e = \sqrt {1 - {9 \over {16}}} $$

$$ \Rightarrow e = {{\sqrt 7 } \over 4}$$

Now, radius of this circle $= {a^2} = 16$

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

Now equation of circle is

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

${x^2} + y{}^2 - 6y - 7 = 0$

Solutions · 1