# The equation of the circle passing through the foci of the ellipse $$x^2\over 16$$ + $$y^2\over 9$$ = 1 and having center at (0, 3) is

## Solution :

Given the equation of ellipse is $$x^2\over 16$$ + $$y^2\over 9$$ = 1

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

$$\therefore$$ foci is ($$\pm ae$$, 0) = ($$\pm\sqrt{7}$$, 0)

$$\therefore$$ Radius of the circle, r = $$\sqrt{(ae)^2+b^2}$$

r = $$\sqrt{7+9}$$ = $$\sqrt{16}$$ = 4

Now, equation of the circle is $$(x-0)^2 + (y-3)^2$$ = 16

$$\therefore$$ $$x^2 + y^2 – 6y – 7$$ = 0

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