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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