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

Given,

Common tangent to the curves $y^2=4x$ and ${\left(x-4\right)}^2+y^2=16$ touch the curves at the points $P$ and $Q$

Let, the equation of tangent be, 

$y=mx+\frac{a}{m}$, where $a=1$ is focus and $m$ is slope,

So, equation of tangent to parabola will be, $y=mx+\frac{1}{m}$

Also line is tangent to circle,

So, using perpendicular distance formula from centre $\left(4, 0\right)$to tangent which will be radius $r=4$, we get, 

$\left|\frac{4m+\frac{1}{m}}{m^2+1}\right|=4$

$\Rightarrow 16m^2+\frac{1}{m^2}+8=16m^2+16$

 $\Rightarrow m^2=\frac{1}{8}$

$\Rightarrow m=\pm \frac{1}{2\sqrt{2}}$

Now if $m=\frac{1}{2\sqrt{2}}$ then $P=\left(\frac{a}{m^2}, \frac{2a}{m}\right)=\left(8, 4\sqrt{2}\right)$

Now equation of tangent will be, $x-2\sqrt{2}y+8=0$

And $Q$ will be foot of perpendicular from $\left(4, 0\right)$ on tangent$x-2\sqrt{2}y+8=0$ ,

So, using foot of perpendicular formula  we get,

$\frac{x-4}{1}=\frac{y-0}{-2\sqrt{2}}=-\frac{4-0+8}{1^2+{\left(2\sqrt{2}\right)}^2}$

$\Rightarrow \left(x,y\right)=\left(\frac{8}{3}, \frac{8\sqrt{2}}{3}\right)$

Hence, by distance formula we get,

${\left(PQ\right)}^2={\left(\frac{16}{3}\right)}^2+{\left(\frac{4\sqrt{2}}{3}\right)}^2$

$\Rightarrow {\left(PQ\right)}^2=\frac{256+32}{9}=32$