JEE Main · 2004 · 24 Apr 2004 · Offline · Parabola

Let the moving circle have centre $(h,k)$. Touching the $x$-axis makes its radius equal to $k$, and passing through $A(p,q)$ gives $\sqrt{(h-p{)}^2+(k-q{)}^2}=k$, which squares to $(h-p{)}^2=2qk-q^2$. If $(x,y)$ is the far end of the diameter through $A$, then $A$ is the midpoint of $A$ and $(x,y)$ reversed, so $h=\frac{x+p}{2}$ and $k=\frac{y+q}{2}$. Substituting, ${\left(\frac{x-p}{2}\right)}^2=2q\cdot \frac{y+q}{2}-q^2$, which simplifies to $(x-p{)}^2=4qy$.