$\begin{array}{1 1}(A)\;4\\(B)\;3 \\(C)\;\sqrt {\large\frac{1}{2}} \\(D)\;\large\frac{7}{2} \end{array}$

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For ellipse $\large\frac{x^2}{4^2}+\frac{y^2}{3^2}$$=1$

$a=4\;b=3$

=>$e =\sqrt {1-\bigg(\large\frac{3}{4} \bigg)^2}$

$\qquad= \sqrt {\large\frac{7}{4}}$

Foci are $(\sqrt 7,0)$ and $(-\sqrt 7,0)$

Centre of circle is at $(0,3)$ and it passes through radius of circle

$\qquad= \sqrt{(7)^2+(3)^2}$

$\qquad= 4$

Hence A is the correct answer.

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