Let S_{1}: x^{2}+y^{2}=9 and S_{2}:(x-2)^{2}+ | vedclass

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(C) Given circles are $S_{1}: x^{2}+y^{2}=3^{2}$ with center $A(0,0)$ and radius $r_{1}=3$,and $S_{2}: (x-2)^{2}+y^{2}=1^{2}$ with center $B(2,0)$ and

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

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(C) Given circles are $S_{1}: x^{2}+y^{2}=3^{2}$ with center $A(0,0)$ and radius $r_{1}=3$,and $S_{2}: (x-2)^{2}+y^{2}=1^{2}$ with center $B(2,0)$ and radius $r_{2}=1$.Let the variable circle $S$ have center $P(x,y)$ and radius $r$.Since $S$ touches $S_{1}$ internally,the distance $PA = r_{1} - r = 3 - r$.Since $S$ touches $S_{2}$ externally,the distance $PB = r_{2} + r = 1 + r$.Adding these two equations,we get $PA + PB = (3 - r) + (1 + r) = 4$.Since $PA + PB = 4$ and the distance $AB = 2$,the locus of $P$ is an ellipse with foci at $A(0,0)$ and $B(2,0)$ and major axis length $2a = 4$,so $a = 2$.The center of the ellipse is the midpoint of $AB$,which is $(1,0)$.The distance between foci is $2ae = AB = 2$,so $2(2)e = 2$,which gives $e = \frac{1}{2}$.Then $b^{2} = a^{2}(1 - e^{2}) = 4(1 - \frac{1}{4}) = 3$.The equation of the ellipse is $\frac{(x-1)^{2}}{4} + \frac{y^{2}}{3} = 1$.Checking the options,for $x=2$,$\frac{(2-1)^{2}}{4} + \frac{y^{2}}{3} = 1$ $\Rightarrow \frac{1}{4} + \frac{y^{2}}{3} = 1$ $\Rightarrow \frac{y^{2}}{3} = \frac{3}{4}$ $\Rightarrow y^{2} = \frac{9}{4}$ $\Rightarrow y = \pm \frac{3}{2}$.Thus,the locus passes through $\left(2, \pm \frac{3}{2}\right)$.

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

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