Solve this following


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:


  1. $(0, \pm \sqrt{3})$

  2. $\left(\frac{1}{2}, \pm \frac{\sqrt{5}}{2}\right)$

  3. $\left(2, \pm \frac{3}{2}\right)$

  4. $(1, \pm 2)$

Correct Option: , 3


$\mathrm{S}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}$


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

$\therefore$ given circle are touching internally

Let a veriable circle with centre $P$ and radius $\mathrm{r}$

$\Rightarrow \mathrm{PA}=\mathrm{r}_{1}-\mathrm{r}$ and $\mathrm{PB}=\mathrm{r}_{2}+\mathrm{r}$

$\Rightarrow \mathrm{PA}+\mathrm{PB}=\mathrm{r}_{1}+\mathrm{r}_{2}$

$\Rightarrow \mathrm{PA}+\mathrm{PB}=4 \quad(>\mathrm{AB})$

$\Rightarrow$ Locus of $\mathrm{P}$ is an ellipse with foci at $\mathrm{A}(0,0)$

and $\mathrm{B}(2,0)$ and length of major axis is $2 \mathrm{a}=4$


$\Rightarrow$ 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 $\left(2, \pm \frac{3}{2}\right)$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now