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 _______.

The standard right-opening parabola $$y^{2}=4ax$$ has parameter $$a=1$$, so its equation is $$y^{2}=4x$$.

Slope form of a tangent to this parabola:
For slope $$m$$ the tangent is $$y=mx+\frac{a}{m}=mx+\frac{1}{m}$$ $$-(1)$$.

The given circle is $$(x-4)^{2}+y^{2}=16$$, centre at $$(4,0)$$, radius $$4$$.

Line $$-(1)$$ is also tangent to the circle. Substitute $$y=mx+\frac{1}{m}$$ into the circle:

$$(x-4)^{2}+\left(mx+\frac{1}{m}\right)^{2}=16$$
$$\bigl(x^{2}-8x+16\bigr)+\bigl(m^{2}x^{2}+2x+\frac{1}{m^{2}}\bigr)=16$$
$$(1+m^{2})x^{2}-6x+\frac{1}{m^{2}}=0$$ $$-(2)$$

For tangency the quadratic $$-(2)$$ must have equal roots, so its discriminant is zero:

$$(-6)^{2}-4(1+m^{2})\frac{1}{m^{2}}=0$$
$$36-4\frac{1+m^{2}}{m^{2}}=0$$
Divide by $$4$$ and multiply by $$m^{2}$$: $$9m^{2}-(1+m^{2})=0$$
$$8m^{2}=1 \;\Longrightarrow\; m^{2}=\frac{1}{8}$$ $$-(3)$$

Thus the two common tangents have slopes $$m=\pm\frac{1}{2\sqrt{2}}$$. (The geometry is symmetric about the x-axis; we work with the positive slope, the other gives the same length.)

Point of contact on the parabola (P)
Insert $$m^{2}=\frac{1}{8}$$ into the intersection of line $$-(1)$$ with $$y^{2}=4x$$:

$$\left(mx+\frac{1}{m}\right)^{2}=4x$$
$$m^{2}x^{2}+2x+\frac{1}{m^{2}}-4x=0$$
$$m^{2}x^{2}-2x+\frac{1}{m^{2}}=0$$

This quadratic also has a double root (tangency), giving $$x_P=\frac{1}{m^{2}}=8, \qquad y_P=mx_P+\frac{1}{m}=2\sqrt{2}+2\sqrt{2}=4\sqrt{2}.$$ Hence $$P\,(8,\,4\sqrt{2})$$.

Point of contact on the circle (Q)
The repeated root of $$-(2)$$ gives the x-coordinate:

$$x_Q=\frac{6}{2(1+m^{2})}=\frac{3}{1+m^{2}}=\frac{3}{1+\frac{1}{8}}=\frac{3}{\frac{9}{8}}=\frac{8}{3}.$$

Then $$y_Q=mx_Q+\frac{1}{m}=\frac{1}{2\sqrt{2}}\cdot\frac{8}{3}+2\sqrt{2}=\frac{4}{3\sqrt{2}}+2\sqrt{2}=\frac{16}{3\sqrt{2}}.$$ Thus $$Q\!\left(\frac{8}{3},\,\frac{16}{3\sqrt{2}}\right).$$

Distance $$PQ$$
$$\Delta x = 8-\frac{8}{3}=\frac{16}{3},\qquad \Delta y = 4\sqrt{2}-\frac{16}{3\sqrt{2}}=\frac{8}{3\sqrt{2}}.$$

Therefore

$$PQ^{2}=(\Delta x)^{2}+(\Delta y)^{2}$$
$$=\left(\frac{16}{3}\right)^{2}+\left(\frac{8}{3\sqrt{2}}\right)^{2}$$
$$=\frac{256}{9}+\frac{64}{18}$$
$$=\frac{256}{9}+\frac{32}{9}=\frac{288}{9}=32.$$

Hence $$PQ^{2}=32$$.

The same value is obtained for the tangent with negative slope, so the answer is unique.

Final Answer : 32