JEE Main 2021 (Online) 18th March Evening ShiftQuestion 1
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 :
Ellipse
Solution
Concept. The two touching conditions make the sum of the distances of the centre from $(0,0)$ and $(2,0)$ constant, so the locus is an ellipse with these foci. Formulas used. (1) Internal touching: distance of centres $=R-r$; external: $=R+r$(2) $PF_1+PF_2=2a>F_1F_2$: ellipse with foci $F_1,F_2$, $b^2=a^2-c^2$Step 1 - Conditions (1). Let $S$ have centre $C$ and radius $r$. With $S_1$ (centre $O$, radius 3): $OC=3-r$. With $S_2$ (centre $(2,0)$, radius 1): $C$ to $(2,0)$ $=1+r$. Step 2 - Add. $OC+C(2,0)=4>2$, so by (2) the locus is an ellipse with foci $(0,0),(2,0)$, $2a=4$. Step 3 - Its equation. Centre $(1,0)$, $a=2$, $c=1$, $b^2=3$: $$\frac{(x-1)^2}{4}+\frac{y^2}{3}=1$$ Step 4 - Check options. $\left(2,\pm\frac32\right)$: $\frac14+\frac34=1$ ✓. $(0,\pm\sqrt3)$: $\frac14+1\ne1$. $(1,\pm2)$: $\frac43\ne1$. $\left(\frac12,\pm\frac{\sqrt5}2\right)$: $\frac1{16}+\frac5{12}\ne1$.(At $C(2,\frac32)$: $r=3-\frac52=\frac12$, and the distance to $(2,0)$ is $\frac32=1+\frac12$.) Answer: A