question_answer

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 :     AIEEE Solevd Paper-2013

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) \[{{x}^{2}}+{{y}^{2}}-6y+5=0\]

Correct Answer: A

Solution :

Co-ordinate of Foci are (ae, 0); (−ae, 0) \[e=\sqrt{1-\frac{9}{16}}\] \[\Rightarrow \] \[e=\frac{\sqrt{7}}{4}\] Co-ordinate of Foci are \[(\sqrt{7},\,0);\,\,(-\sqrt{7},\,0)\] \[R=\sqrt{7+9}=4\] \[{{(x-0)}^{2}}+{{(y-3)}^{2}}={{4}^{2}}\] \[{{x}^{2}}+{{y}^{2}}-6y+9=16\] \[\therefore \] \[{{x}^{2}}+{{y}^{2}}-6y-7=0\]