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

Answer 1

The correct option is B $(2,\pm \frac{3}{2})$

$S_1:x^2+y^2=9S_2:{(x-2)}^2+y^2=1$
The centre and radius are
$C_1:(0,0),r_1=3C_2:(2,0),r_2=1$
Let centre of variable circle be $C_3(h,k)$ and radius be $r$


$C_3{C}_1=3-r{C}_2{C}_3=1+r{C}_3{C}_1+C_2{C}_3=4\Rightarrow \sqrt{h^2+k^2}+\sqrt{(h-2{)}^2+k^2}=4\Rightarrow \sqrt{(h-2{)}^2+k^2}=4-\sqrt{h^2+k^2}\Rightarrow (h-2{)}^2+k^2=16+h^2+k^2-8\sqrt{h^2+k^2}\Rightarrow -4h+4=16-8\sqrt{h^2+k^2}\Rightarrow h+3=2\sqrt{h^2+k^2}\Rightarrow h^2+6h+9=4h^2+4k^2\Rightarrow 3(h-1{)}^2+4k^2=12\Rightarrow k=\pm \sqrt{3}\sqrt{1-{\left(\frac{h-1}{2}\right)}^2}$
From the given options
$(2,\pm \frac{3}{2})$ satisfies it.