The equation of the circle passing through the foci of the ellipse, and having centre at $(0, 3)$ is

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

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

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x^{2} + y^{2} - 6 y - 5 = 0

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x^{2} + y^{2} - 6 y + 5 = 0

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Solution

The correct option is A $x^{2} + y^{2} - 6 y - 7 = 0$
Co-ordinates of foci $(a e, 0) (- a e, 0)$
$e = \sqrt{1 - \frac{9}{16}} e = \sqrt{7}$
Co-ordinate of foci are $(\sqrt{7}, 0), (- \sqrt{7}, 0) R = \sqrt{7 + 9} = 4 {(x - 0)}^{2} + {(y - 3)}^{2} = 4^{2} x^{2} + y^{2} - 6 y + 9 = 16 x^{2} + y^{2} - 6 y - 7 = 0$

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The diameters of a circle are along $2 x + y - 7 = 0$ and $x + 3 y - 11 = 0$ Then the equation of this circle, which also passes through (5,7) is

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