Question

Find radii of smallest and largest circle passing through the point $(\sqrt{3},\sqrt{2})$ and touching the circle$x^2+y^2-2\sqrt{2}y-2=0are{r}_{1}and{r}_{{}_2}$ respectively.

Answer 1

Position of $(\sqrt{3},\sqrt{2})$ with respect to $x^2+y^2-2\sqrt{2}y-2=0$

$s_1=(\sqrt{3}{)}^2+(\sqrt{2}{)}^2-2\sqrt{2}\left(\sqrt{2}\right)-2$

$=3+2-6=-1$

$s_1<0$

$\Rightarrow (\sqrt{3},\sqrt{2})$ lies inside the circle centre of given circle is $(0,\sqrt{2})$

Radius $=\sqrt{(\sqrt{2}{)}^2+2}=2$

$\Rightarrow CA=2$

$CB=2$

Circle with passing through P and touching the given circle

For smallest circle

PA should be taken as diameter

$PA=2r_1=CA-CP$

$2r_1=2-\sqrt{3}$

$\Rightarrow r_1=1-\frac{\sqrt{3}}{2}$

For largest circle PB should be taken as diameter

$PB=2r_2=CB+CP$

$=2+\sqrt{3}$

$\Rightarrow r_2=1+\frac{\sqrt{3}}{2}$.