Question

The radius of the circle passing through the foci of the ellipse $\frac{x^2}{4}+\frac{4y^2}{7}=1$ and having its centre at $(\frac{1}{2},2)$ is

• A. $\sqrt{5}$
• B. $3$
• C. $\sqrt{12}$
• D. $\frac{7}{2}$

Answer 1

Answer: (A)

$\sqrt{5}$

Given equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{\frac{7}{4}}=1$
Here, $a^2=4$ and $b^2=\frac{7}{4}$
Since $b^2=a^2(1-e^2)$
Thus $\frac{7}{4}=4(1-e^2)$
$\Rightarrow e2=1-\frac{7}{16}=\frac{9}{16}\Rightarrow e=\frac{3}{4}$,
Thus, the foci are $(\pm \frac{3}{2},0)$.
The radius of the required circle
$=\sqrt{{(\frac{3}{2}-\frac{1}{2})}^2+(0-2{)}^2}$
$=\sqrt{5}$