The radius of the circle having its centre at | vedclass

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(c) The co-ordinates of foci are $( \pm ae,\,0)$. Here $a = 4,\,\,b = 3$

$	herefore$  ${b^2} = {a^2}(1 - {e^2})$ ==> $9 = 16(1 - {e^2})$&n

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(c) The co-ordinates of foci are $( \pm ae,\,0)$. Here $a = 4,\,\,b = 3$

$	herefore$  ${b^2} = {a^2}(1 - {e^2})$ ==> $9 = 16(1 - {e^2})$  ==> $\frac{9}{{16}} = 1 - {e^2}$

==> $e = \pm \sqrt {\frac{7}{4}} $; $	herefore$ Points are $( \pm \sqrt 7 ,0)$.

$	herefore$  Radius $ = \sqrt {{{(\sqrt 7 - 0)}^2} + {{(0 - 3)}^2}} = \sqrt {7 + 9} = \sqrt {16} = 4$.

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

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