Question:

Let a common tangent to $y^2=4x$ and $(x-4)^2+y^2=16$ touch the curves at $P$ and $Q$. Then $(PQ)^2$ is equal to ________.

Step-by-Step Solution

Key Concept: Tangent to parabola: $y=mx+\frac{1}{m}$. Tangency to circle $(x-4)^2+y^2=16$: distance from $(4,0)$ to line $=4$. Solve for $m$.

$(PQ)^2=32$.

Correct Answer: 32