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. $(\frac{1}{2}, \pm \frac{\sqrt{8}}{2})$
• B. $(0, \pm \sqrt{3})$
• C. $(2, \pm \frac{3}{2})$
• D. $(1, \pm 2)$

Answer 1

Answer: (C)

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

Answer 2

Let $O_1=(0,0)$ and $R_1=3$ be the center and radius of $S_1$. Let $O_2=(2,0)$ and $R_2=1$ be the center and radius of $S_2$. Let $O=(x,y)$ and $r$ be the center and radius of the variable circle $S$.

Since $S$ touches $S_1$ internally, the distance between their centers is $d(O_1, O) = R_1 - r$. $\sqrt{x^2 + y^2} = 3 - r \quad (*)$ Since $S$ touches $S_2$ externally, the distance between their centers is $d(O_2, O) = R_2 + r$. $\sqrt{(x-2)^2 + y^2} = 1 + r \quad (**)$ From (*), $r = 3 - \sqrt{x^2 + y^2}$. Substituting this into (**): $\sqrt{(x-2)^2 + y^2} = 1 + (3 - \sqrt{x^2 + y^2})$ $\sqrt{(x-2)^2 + y^2} = 4 - \sqrt{x^2 + y^2}$ $\sqrt{x^2 + y^2} + \sqrt{(x-2)^2 + y^2} = 4$ This is the definition of an ellipse with foci at $F_1=(0,0)$ and $F_2=(2,0)$. The sum of the distances is $2a=4$, so $a=2$. The distance between the foci is $2c = \sqrt{(2-0)^2 + (0-0)^2} = 2$, so $c=1$. Using $a^2 = b^2 + c^2$, we get $2^2 = b^2 + 1^2$, so $4 = b^2 + 1$, which means $b^2 = 3$. The center of the ellipse is the midpoint of the foci: $(\frac{0+2}{2}, \frac{0+0}{2}) = (1,0)$. The equation of the ellipse is $\frac{(x-1)^2}{a^2} + \frac{y^2}{b^2} = 1$, which is $\frac{(x-1)^2}{4} + \frac{y^2}{3} = 1$.

Now we check the points: For $(2, \pm \frac{3}{2})$: $\frac{(2-1)^2}{4} + \frac{(\pm \frac{3}{2})^2}{3} = \frac{1^2}{4} + \frac{\frac{9}{4}}{3} = \frac{1}{4} + \frac{9}{12} = \frac{1}{4} + \frac{3}{4} = 1$. This point lies on the ellipse.