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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Aicharya · Accepted Answer

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, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$ $	herefore \quad 	ext { Focus }=( \pm a e, 0)= \pm \sqrt{7}, 0$ Since, the circle passes through $\mathrm{P}( \pm \sqrt{7}, 0)$ and centre at $\mathrm{C}(0,3)$. $\begin{aligned} & 	herefore \quad 	ext { Radius }=r c p=\sqrt{(0-7)^2+(3-0)^2} \ & =\sqrt{7+9}=\sqrt{16}=4 \end{aligned}$

The radius of the circle passing through the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ and having…

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