Let S_1: x^2+y^2=9 and S_2:(x-2)^2+y^2=1.

Question

Let $S_1: x^2+y^2=9$ and $S_2:(x-2)^2+y^2=1$. Then the locus of the centre of a variable circle $S$ which touches $S_1$ internally and $S_2$ externally always passes through the points

Accepted Answer

$C_1=(0,0),r_1=3$; $C_2=(2,0),r_2=1$. For the variable circle (radius $r$, centre $P$) touching $S_1$ internally and $S_2$ externally:
$$PC_1=3-r,\qquad PC_2=1+r\ \Longrightarrow\ PC_1+PC_2=4.$$

This is an ellipse with foci $C_1,C_2$, $2a=4\Rightarrow a=2$, $2ae=C_1C_2=2\Rightarrow e=\frac12$, $b^2=a^2(1-e^2)=3$. Centre at midpoint $(1,0)$:
$$\frac{(x-1)^2}4+\frac{y^2}3=1.$$

Checking the options, only $\left(2,\pm\frac32\right)$ satisfies this equation: $\frac14+\frac{9/4}3=\frac14+\frac34=1$. *(Note: the source answer key marks this "A", but the equation derived --- confirmed by the source's own solution --- is satisfied only by option C.)*

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