The radius of the circle passing through the foci of the ellipse x^216+ y^29=1 and having its centre at (0 ,3 ) is

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Accepted Answer

Correct answer: B. (b) Equation of ellipse, x^216+ y^29=1 a^2=16 ⇒a=4 ,b^2=9 ⇒b=3 Here, a>b Now, e= 1- b^2a^2= 1- 916= 74 ∴ Focus = ( ae ,0 )= ( 7 ,0 ) Since, the circle passes through P ( 7 ,0 ) and centre at C (0 ,3 ). ∴ Radius = CP= (0-7 )^2+ (3-0 )^2= 7+9= 16=4

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

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