JEE MAIN - Mathematics (2013 (Offline) - No. 10)
The equation of the circle passing through the foci of the ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$, and having centre at $$(0,3)$$ is :
$${x^2} + {y^2} - 6y - 7 = 0$$
$${x^2} + {y^2} - 6y + 7 = 0$$
$${x^2} + {y^2} - 6y - 5 = 0$$
$${x^2} + {y^2} - 6y + 5 = 0$$
Explanation
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From the given equation of ellipse, we have
$$a = 4,b = 3,e = \sqrt {1 - {9 \over {16}}} $$
$$ \Rightarrow e = {{\sqrt 7 } \over 4}$$
Now, radius of this circle $$ = {a^2} = 16$$
$$ \Rightarrow Focii = \left( { \pm \sqrt 7 ,0} \right)$$
Now equation of circle is
$${\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} = 16$$
$${x^2} + y{}^2 - 6y - 7 = 0$$
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