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 $\left(\frac{1}{2},2\right)$ is

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

Answer 1

Answer: (A)

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