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 $PB=r_2+r$
$\Rightarrow PA+PB=r_1+r_2$
$\Rightarrow PA+PB=4(>AB)$
$\Rightarrow$ Locus of $P$ is an ellipse with foci at $A(0,0)$ and $B(2,0)$ and length of major axis is $2a=4$.
$e=\frac{1}{2}$
$\Rightarrow$ 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 $\left(2,\pm \frac{3}{2}\right)$