Question

Let $S_{1}: x^{2}+y^{2}=9$ and $S_{2}:(x-2)^{2}+y^{2}=1$. Then the locus of center of a variable circle $S$ which touches $S _{1}$ internally and $S _{2}$ externally always passes through the points:

• A. $(0, \pm \sqrt{3})$
• B. $\left(\frac{1}{2}, \pm \frac{\sqrt{5}}{2}\right)$
• C. $\left(2, \pm \frac{3}{2}\right)$
• D. $(1, \pm 2)$

Answer 1

Answer: (C)

$\because c_{1} c_{2}=r_{1}-r_{2}$

$\therefore $ given circle are touching internally
Let a veriable circle with centre $P$ and radius $r$
$\Rightarrow PA = r _{1}- r$ and $PB = r _{2}+ r$
$\Rightarrow PA + PB = r _{1}+ r _{2}$
$\Rightarrow PA + PB =4 \quad(> AB )$
$\Rightarrow $ Locus of $P$ is an ellipse with foci at $A (0,0)$ and $B (2,0)$ and length of major axis is $2 a =4$.
$e =\frac{1}{2}$
$\Rightarrow $ centre is at $(1,0)$ and $b ^{2}= a ^{2}\left(1- e ^{2}\right)=3$ if $x$ -ellipse

$\Rightarrow E : \frac{( x -1)^{2}}{4}+\frac{ y ^{2}}{3}=1$
which is satisfied by $\left(2, \pm \frac{3}{2}\right)$