Ellipse: The radius of the circle passing through the foci of the ellipse… | IIT-JEE 1995 Screening | Yukti Vault

Yukti Vault

IIT-JEE 1995 ScreeningQuestion 1

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

Ellipse

$4$✓ Correct answer$3$$\sqrt{\dfrac12}$$\dfrac72$

Solution

Foci $(\pm\sqrt{16-9},0)=(\pm\sqrt7,0)$. Radius$^2=7+3^2=16$, so $r=4$.

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