JEE PYQ: Circle Question 9
Question 9 - 2021 (18 Mar Shift 2)
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) $(1, \pm 2)$
Show Answer
Answer: (3)
Solution
Let variable circle have center $P$ and radius $r$. $PA = 3 - r$ (internal tangency with $S_1$), $PB = r + 1$ (external tangency with $S_2$). So $PA + PB = 4 > AB = 2$. Locus is an ellipse with foci $A(0,0)$ and $B(2,0)$, $2a = 4$. Center $(1,0)$, $b^2 = 3$. The point $(2, \pm\frac{3}{2})$ lies on this ellipse: $\frac{1}{4} + \frac{9/4}{3} = 1$.
BETA
