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The radius of the circle passing through the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ and having its centre at $(0,3)$ is
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The correct answer is:
$4$
Equation of ellipse, $\frac{x^2}{16}+\frac{y^2}{9}=1$
$a^2=16 \Rightarrow a=4, b^2=9 \Rightarrow b=3$
Here, $a>b$
Now, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
$\therefore \quad \text { Focus }=( \pm a e, 0)= \pm \sqrt{7}, 0$
Since, the circle passes through $\mathrm{P}( \pm \sqrt{7}, 0)$ and centre at $\mathrm{C}(0,3)$.
$\begin{aligned}
& \therefore \quad \text { Radius }=r c p=\sqrt{(0-7)^2+(3-0)^2} \\
& =\sqrt{7+9}=\sqrt{16}=4
\end{aligned}$
$a^2=16 \Rightarrow a=4, b^2=9 \Rightarrow b=3$
Here, $a>b$
Now, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
$\therefore \quad \text { Focus }=( \pm a e, 0)= \pm \sqrt{7}, 0$
Since, the circle passes through $\mathrm{P}( \pm \sqrt{7}, 0)$ and centre at $\mathrm{C}(0,3)$.
$\begin{aligned}
& \therefore \quad \text { Radius }=r c p=\sqrt{(0-7)^2+(3-0)^2} \\
& =\sqrt{7+9}=\sqrt{16}=4
\end{aligned}$
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