Ellipse 1 Question 4

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

(2013 Main)

(a) $x^{2}+y^{2}-6 y-7=0$

(c) $x^{2}+y^{2}-6 y-5=0$

(b) $x^{2}+y^{2}-6 y+7=0$

(d) $x^{2}+y^{2}-6 y+5=0$

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Answer:

Correct Answer: 5. (a)

Solution:

  1. Given equation of ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$

Here, $a=4, b=3, e=\sqrt{1-\frac{9}{16}} \Rightarrow \frac{\sqrt{7}}{4}$

$\therefore$ Foci $=( \pm a e, 0)= \pm 4 \times \frac{\sqrt{7}}{4}, 0=( \pm \sqrt{7}, 0)$

Radius of the circle, $r=\sqrt{(a e)^{2}+b^{2}}$

$ =\sqrt{7+9}=\sqrt{16}=4 $

Now, equation of circle is

$ \begin{aligned} (x-0)^{2}+(y-3)^{2} & =16 \\ \therefore \quad x^{2}+y^{2}-6 y-7 & =0 \end{aligned} $



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