Let a common tangent to the curves y^2 = 4x and (x - 4)^2 + y^2 = 16 touch the curves at t…

Question

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

Answer: 32

Accepted Answer

Tangent of slope $m$ to the parabola

${y}^{2}=4x$ is given by $y=mx+\frac{1}{m}$ and

Tangent of slope $m$ to the circle $(x-4{)}^{2}+{y}^{2}=16$ is given by

$y=m(x-4)\pm 4\sqrt{1+{m}^{2}}$

For common tangent

$\begin{gathered} \frac{1}{m}=-4m\pm 4\sqrt{1+{m}^{2}} \ \Rightarrow {(\frac{1}{m}+4m)}^{2}={(\pm 4\sqrt{1+{m}^{2}})}^{2} \end{gathered}$

On squaring both sides, we get

$\begin{gathered} =\frac{1}{{m}^{2}}+16{m}^{2}+8=16+16{m}^{2} \ \Rightarrow \frac{1}{{m}^{2}}=8\Rightarrow m=\pm \frac{1}{2\sqrt{2}} \end{gathered}$

Then, the point of contact on parabola is $(8,4\sqrt{2})$

Length of tangent $PQ$ from $(8,4\sqrt{2})$ on the circle is

$\begin{gathered} \Rightarrow PQ=\sqrt{(8-4{)}^{2}+(4\sqrt{2}{)}^{2}-16} \ \Rightarrow PQ=\sqrt{16+32-16} \ \Rightarrow PQ=\sqrt{32}\Rightarrow (PQ{)}^{2}=32 \end{gathered}$

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