NIMCET 2015 Question Paper with solutions

We have
Equation of ellipse is $\frac{{\mathit{x}}^2}{16}+\frac{{\mathit{y}}^2}{9}=1$
Here, ${\mathit{a}}^2=16$ and ${\mathit{b}}^2=9$
We know that, foci of ellipse of
$\frac{{\mathit{x}}^2}{{\mathit{a}}^2}+\frac{{\mathit{y}}^2}{{\mathit{b}}^2}=1$ is (±ae,0)
$=\sqrt{16-9}$
$=\sqrt{7}$
∴ Foci of ellipse $\left(\pm \sqrt{7},0\right)$
It is given that, centre of circle is (0, 3) and passes through $\pm \sqrt{7},0$ $\therefore \mathit{r}=\sqrt{{\left(0+\sqrt{7}\right)}^2+{\left(3-0\right)}^2}$
$\sqrt{7+9}$
$=4$
Hence, radius of circle = 4 units