Let the tangents drawn from the origin to the circle, $$x^2 + y^2 - 8x - 4y + 16 = 0$$ touch it at the points A and B. Then $$(AB)^2$$ is equal to

We have the circle $$x^{2}+y^{2}-8x-4y+16=0$$. To recognise its centre and radius, we complete squares.

Re-grouping the $$x$$ and $$y$$ terms,

$$x^{2}-8x+y^{2}-4y+16=0.$$

Inside each bracket we add and subtract the square of half the coefficient:

$$\bigl(x^{2}-8x+16\bigr)+\bigl(y^{2}-4y+4\bigr)+16-16-4=0.$$

This simplifies to

$$\left(x-4\right)^{2}+\left(y-2\right)^{2}-4=0,$$

so in standard form

$$\left(x-4\right)^{2}+\left(y-2\right)^{2}=4.$$

Hence the centre is $$C(4,\,2)$$ and the radius is $$r=2$$.

Let $$O(0,\,0)$$ be the origin. We first find the distance $$OC$$:

$$OC=\sqrt{4^{2}+2^{2}}=\sqrt{16+4}=2\sqrt5.$$

For any external point, the square of the length of the tangent drawn to a circle equals the power of the point, namely

$$\text{(tangent length)}^{2}=OC^{2}-r^{2}.$$

Substituting $$OC^{2}=20$$ and $$r^{2}=4$$ gives

$$OA^{2}=OB^{2}=20-4=16\;\Longrightarrow\;OA=OB=4.$$

The tangents touch the circle at $$A$$ and $$B$$, and the radii $$CA$$ and $$CB$$ are perpendicular to the tangents. Thus each of the triangles $$\triangle OCA$$ and $$\triangle OCB$$ is right-angled at $$A$$ and $$B$$ respectively.

Let $$\alpha=\angle OCA$$. In $$\triangle OCA$$ we know three sides: $$OC=2\sqrt5,\; CA=r=2,\; OA=4.$$ Applying the Cosine Rule (stated: in any triangle $$a^{2}=b^{2}+c^{2}-2bc\cos A$$), with side $$OA$$ opposite angle $$\alpha$$, we get

$$OA^{2}=OC^{2}+CA^{2}-2\,(OC)\,(CA)\cos\alpha.$$

Substituting the lengths,

$$16=20+4-2\,(2\sqrt5)\,(2)\cos\alpha.$$

Simplifying,

$$16=24-8\sqrt5\,\cos\alpha,$$

$$8\sqrt5\,\cos\alpha=24-16=8,$$

$$\cos\alpha=\dfrac{8}{8\sqrt5}=\dfrac1{\sqrt5}.$$

Therefore

$$\sin\alpha=\sqrt{1-\cos^{2}\alpha}=\sqrt{1-\dfrac1{5}}=\sqrt{\dfrac45}=\dfrac{2}{\sqrt5}.$$

The angle between the two radii $$CA$$ and $$CB$$ is $$\angle ACB=2\alpha$$ because the triangles $$\triangle OCA$$ and $$\triangle OCB$$ are mirror images about $$OC$$. For a circle, the length of a chord subtending an angle $$\theta$$ at the centre is

$$\text{chord}=2r\sin\frac{\theta}{2}.$$

Here the chord is $$AB$$, the subtended angle is $$\theta=2\alpha$$, so

$$AB = 2r\sin\frac{2\alpha}{2}=2r\sin\alpha.$$

Substituting $$r=2$$ and $$\sin\alpha=\dfrac{2}{\sqrt5}$$,

$$AB = 2\,(2)\,\dfrac{2}{\sqrt5}= \dfrac{8}{\sqrt5}.$$

Finally, the required quantity is

$$\bigl(AB\bigr)^{2}= \left(\dfrac{8}{\sqrt5}\right)^{2}= \dfrac{64}{5}.$$

Hence, the correct answer is Option C.