Let $$P$$ be the point on the parabola, $$y^2 = 8x$$ which is at a minimum distance from the center $$C$$ of the circle, $$x^2 + (y+6)^2 = 1$$. Then the equation of the circle, passing through $$C$$ and having its center at $$P$$ is

We have the parabola $$y^{2}=8x$$. A general point on this parabola can be written as $$P(x,y)$$ with the relation $$x=\dfrac{y^{2}}{8}$$.

The circle given in the question is $$x^{2}+(y+6)^{2}=1$$. Its centre is clearly obtained by comparing with $$(x-h)^{2}+(y-k)^{2}=r^{2}$$; we get $$C(0,-6)$$.

We require the point $$P$$ on the parabola that is at the minimum distance from the fixed point $$C$$. Let us denote that (squared) distance by $$D^{2}$$. Using the distance‐formula,

$$D^{2}=(x-0)^{2}+(y+6)^{2}=x^{2}+(y+6)^{2}.$$

Because $$x=\dfrac{y^{2}}{8}$$, we substitute to obtain a single‐variable function:

$$D^{2}(y)=\left(\dfrac{y^{2}}{8}\right)^{2}+(y+6)^{2}=\dfrac{y^{4}}{64}+(y+6)^{2}.$$

To find the minimum we differentiate with respect to $$y$$ and equate to zero. The rule used is “minimum or maximum occurs where the derivative vanishes.”

First derivative:

$$\frac{d}{dy}\!\left[D^{2}(y)\right] =\frac{d}{dy}\left(\dfrac{y^{4}}{64}\right)+\frac{d}{dy}(y+6)^{2} =\dfrac{4y^{3}}{64}+2(y+6) =\dfrac{y^{3}}{16}+2y+12.$$

Setting the derivative to zero gives the cubic equation

$$\dfrac{y^{3}}{16}+2y+12=0 \;\;\Longrightarrow\;\; y^{3}+32y+192=0.$$ Multiplying through by 16 clears the denominator.

We now solve $$y^{3}+32y+192=0$$. A quick integral root test shows $$y=-4$$ is a root because

$$(-4)^{3}+32(-4)+192=-64-128+192=0.$$

Dividing the cubic by $$(y+4)$$ gives the quadratic remainder $$y^{2}-4y+48$$ whose discriminant is negative $$(\Delta=-176<0)$$, so $$y=-4$$ is the only real solution. Hence at the required point we have

$$y_{P}=-4.$$

Using $$x=\dfrac{y^{2}}{8}$$ we get

$$x_{P}=\dfrac{(-4)^{2}}{8}=\dfrac{16}{8}=2.$$

Thus the point of minimum distance is $$P(2,-4).$$

The distance $$PC$$ itself (which will become the radius of the required circle) is found using the distance formula again:

$$PC=\sqrt{(2-0)^{2}+(-4+6)^{2}} =\sqrt{2^{2}+2^{2}} =\sqrt{4+4} =\sqrt{8} =2\sqrt{2}.$$

Now we construct the circle whose centre is $$P(2,-4)$$ and which passes through $$C(0,-6)$$; thus its radius is $$2\sqrt{2}$$. The standard form $$(x-h)^{2}+(y-k)^{2}=r^{2}$$ therefore reads

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

We expand to match the options:

$$\begin{aligned} (x-2)^{2}+(y+4)^{2}&=8\\[4pt] \bigl(x^{2}-4x+4\bigr)+\bigl(y^{2}+8y+16\bigr)&=8\\[4pt] x^{2}+y^{2}-4x+8y+20&=8\\[4pt] x^{2}+y^{2}-4x+8y+12&=0. \end{aligned}$$

Comparing with the given alternatives we see that this is exactly Option C.

Hence, the correct answer is Option C.