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# The radius of the circle passing through the foci of the ellipse $\large\frac{x^2}{16}+\frac{y^2}{9}$$=1 and having its center at (0,3) is \begin{array}{1 1}(A)\;4\\(B)\;3 \\(C)\;\sqrt {\large\frac{1}{2}} \\(D)\;\large\frac{7}{2} \end{array} Can you answer this question? ## 1 Answer 0 votes For ellipse \large\frac{x^2}{4^2}+\frac{y^2}{3^2}$$=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.