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More # 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 A.${x^2} + {y^2} - 6y - 7 = 0$ B.${x^2} + {y^2} - 6y + 7 = 0$ C.${x^2} + {y^2} - 6y - 5 = 0$ D.${x^2} + {y^2} - 6y + 5 = 0$ Verified
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Hint-In this question we have to find the equation of the circle passing through the foci of the ellipse $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1$ but one must know briefly that ellipse is a plane curve surrounding two focal points and ellipse is determined by its foci as such, it generalises a circle but if you can use the lengths of the major and the minor axes to find its coordinates.

Coordinates of Foci are $( \pm ae,0)$
$a = 4,b = 3,e = \sqrt {1 - \dfrac{9}{{16}}} = \dfrac{{\sqrt 7 }}{4}$
Radius of the circle = $\sqrt {{{(ae)}^2} + {3^2}}$ =$\sqrt {16} = 4$
${\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} = 16 \\ \Rightarrow {x^2} + {y^2} - 6y - 7 = 0 \\$