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 __________.

Tangent of slope $m$ to the parabola <br/><br/>$y^2=4 x$ is given by $y=m x+\frac{1}{m}$ and <br/><br/>Tangent of slope $m$ to the circle $(x-4)^2+y^2=16$ is given by <br/><br/>$y=m(x-4) \pm 4 \sqrt{1+m^2}$ <br/><br/>For common tangent <br/><br/>$$ \begin{aligned} & \frac{1}{m}=-4 m \pm 4 \sqrt{1+m^2} \\\\ & \Rightarrow \left(\frac{1}{m}+4 m\right)^2=\left( \pm 4 \sqrt{1+m^2}\right)^2 \end{aligned} $$ <br/><br/>On squaring both sides, we get <br/><br/>$$ \begin{aligned} & =\frac{1}{m^2}+16 m^2+8=16+16 m^2 \\\\ & \Rightarrow \frac{1}{m^2} =8 \Rightarrow m= \pm \frac{1}{2 \sqrt{2}} \end{aligned} $$ <br/><br/>Then, the point of contact on parabola is $(8,4 \sqrt{2})$ <br/><br/>Length of tangent $P Q$ from $(8,4 \sqrt{2})$ on the circle is <br/><br/>$$ \begin{array}{ll} &\Rightarrow P Q=\sqrt{(8-4)^2+(4 \sqrt{2})^2-16} \\\\ &\Rightarrow P Q=\sqrt{16+32-16} \\\\ &\Rightarrow P Q=\sqrt{32} \Rightarrow(P Q)^2=32 \end{array} $$