Question

The centre of the circle of minimum radius passing through $(1,3)$ and touching the circle $2x^2+2y^2-9x-2y+5=0$ is

• A. $\left(\frac{5}{2},\frac{5}{4}\right)$
• B. $\left(\frac{5}{4},\frac{5}{2}\right)$
• C. $\left(\frac{5}{4},-\frac{1}{2}\right)$
• D. $\left(\frac{5}{4},\frac{1}{2}\right)$

Answer 1

The correct option is B $\left(\frac{5}{4},\frac{5}{2}\right)$

The center and radius of the given circle are
$C_1(\frac{9}{4},\frac{1}{2})$ and $r_1=\sqrt{\frac{81}{16}+\frac{1}{4}-\frac{5}{2}}=\frac{3\sqrt{5}}{4}$
The smallest circle touching the circle and passing through $A(1,3)$
should have its center $C_2$ on ${AC}_1$ such that
$AB=2{AC}_2$, where $B$ is the point of intersection of ${AC}_1$ and the given circle ${2r}_2={AC}_1-{BC}_1$
$\Rightarrow 2r_2=\sqrt{{(\frac{9}{4}-1)}^2+{(\frac{1}{2}-3)}^2}-\frac{3\sqrt{5}}{4}=\frac{2\sqrt{5}}{4}$
$\therefore C_2(x,y)$ dividse ${AC}_1$ in the ratio ${AC}_2:C_2{C}_1=\frac{\sqrt{5}}{4}:\frac{4\sqrt{5}}{4}=1:4$
$\therefore x=\frac{1.\frac{9}{4}+4.1}{5}$ and $y=\frac{1.\frac{1}{2}+4.3}{5}$
$i.e.,C_2(\frac{5}{4},\frac{5}{2})$