Let $P$ be a point on a circle $(x-2)^2 + (y-2)^2 = 4$. Now let $Q$ be a point on the same line with $O$, the origin, $P$, and also on the first quadrant. If $Q$ satisfies $\overline{OP}\times\overline{OQ}=6$, the trace of $Q$ is a circle, $(x-3)^2 + (y-3)^2 = 9$.
Is there a generalized theorem or an easy explanation?
You are inverting in the circle $x^2+y^2=6$ (see inversive geometry on Wikipedia). The inverse of a circleline is a circleline. Since your starting circle doesn't pass through the origin, it ends up as a circle.