JEE Main · 2005 · 08 May 2005 · Offline · Parabola

Let the moving circle have centre $(\alpha ,\beta )$; touching the $x$-axis makes its radius $|\beta |$. The fixed circle has centre $(0,3)$ and radius $2$, so for the two to touch, the distance between centres equals the sum of the radii: $\sqrt{{\alpha }^2+(\beta -3{)}^2}=\beta +2$. Squaring, ${\alpha }^2+{\beta }^2-6\beta +9={\beta }^2+4\beta +4$, which simplifies to ${\alpha }^2=10\beta -5$. Replacing $(\alpha ,\beta )$ by $(x,y)$, the locus of the centre is $x^2=10(y-\frac{1}{2})$, a parabola.