The circle passing through foci of frac{{{x^2}}}{{16}} + frac{{{y^2}}}{9} = 1 and having centr

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10-2. Parabola, Ellipse, Hyperbola

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

A

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

B

$\;{x^2} + {y^2} - 6y + 7 = 0$

C

$\;{x^2} + {y^2} - 6y - 5 = 0$

D

$\;{x^2} + {y^2} - 6y + 5 = 0$

(JEE MAIN-2013)

Solution

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

Foci is $(\pm \text { ae }, 0)$ $\Rightarrow \quad(\pm \sqrt{7}, 0)$

$r=\sqrt{(a e)^{2}+b^{2}}$

$\sqrt{7+9}$

$=4$

Now equation of circle is $(x-0)^{2}+(y-3)^{2}=16$

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

Standard 11

Mathematics

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