If a point $$P(0, -2)$$ and $$Q$$ is any point on the circle, $$x^2 + y^2 - 5x - y + 5 = 0$$, then the maximum value of $$(PQ)^2$$ is:

We have point $$P(0,-2)$$ fixed in the plane, and a variable point $$Q(x,y)$$ that lies on the circle

$$x^2 + y^2 - 5x - y + 5 = 0.$$

Our aim is to obtain the maximum possible value of $$(PQ)^2,$$ the square of the distance between $$P$$ and $$Q.$$

First we rewrite the circle in centre-radius form. We take the given equation

$$x^2 + y^2 - 5x - y + 5 = 0$$

and complete the square in both $$x$$ and $$y.$$

Group the $$x$$-terms and the $$y$$-terms:

$$x^2 - 5x \;+\; y^2 - y \;+\; 5 = 0.$$

For the $$x$$-part $$x^2 - 5x$$ we add and subtract $$\left(\dfrac{5}{2}\right)^2 = \dfrac{25}{4},$$ and for the $$y$$-part $$y^2 - y$$ we add and subtract $$\left(\dfrac12\right)^2 = \dfrac14.$$ Writing each addition and subtraction explicitly, we get

$$\bigl[x^2 - 5x + \dfrac{25}{4}\bigr] \;+\; \bigl[y^2 - y + \dfrac14\bigr] \;+\; 5 \;-\; \dfrac{25}{4} \;-\; \dfrac14 \;=\; 0.$$

Now the bracketed expressions are perfect squares:

$$\bigl(x - \dfrac{5}{2}\bigr)^2 + \bigl(y - \dfrac12\bigr)^2 + 5 - \dfrac{25}{4} - \dfrac14 = 0.$$

Simplify the constant term:

$$5 - \dfrac{25}{4} - \dfrac14 = 5 - \dfrac{26}{4} = 5 - 6.5 = -1.5 = -\dfrac32.$$

So the circle becomes

$$\bigl(x - \dfrac{5}{2}\bigr)^2 + \bigl(y - \dfrac12\bigr)^2 = \dfrac32.$$

Hence the centre of the circle is

$$C\left(\dfrac{5}{2},\; \dfrac12\right)$$

and its radius is

$$r = \sqrt{\dfrac32}.$$

Next we find the distance from the fixed point $$P(0,-2)$$ to the centre $$C$$. By the distance formula,

$$PC = \sqrt{\Bigl(\dfrac{5}{2} - 0\Bigr)^2 + \Bigl(\dfrac12 - (-2)\Bigr)^2}.$$

Compute each squared difference:

$$\Bigl(\dfrac{5}{2}\Bigr)^2 = \dfrac{25}{4}, \qquad \Bigl(\dfrac12 + 2\Bigr) = \dfrac12 + \dfrac{4}{2} = \dfrac{5}{2}, \qquad \Bigl(\dfrac{5}{2}\Bigr)^2 = \dfrac{25}{4}.$$

Add the two squares:

$$PC^2 = \dfrac{25}{4} + \dfrac{25}{4} = \dfrac{50}{4} = \dfrac{25}{2}.$$

Hence

$$PC = \sqrt{\dfrac{25}{2}} = \dfrac{5}{\sqrt2}.$$

A standard geometric fact says: for a fixed external point $$P$$ and a circle with centre $$C$$ and radius $$r$$, the farthest point $$Q$$ on the circle lies on the line $$PC$$ extended, and the maximum distance is

$$PQ_{\text{max}} = PC + r.$$

Substituting $$PC = \dfrac{5}{\sqrt2}$$ and $$r = \sqrt{\dfrac32},$$ we have

$$PQ_{\text{max}} = \dfrac{5}{\sqrt2} + \sqrt{\dfrac32} = \dfrac{1}{\sqrt2}\bigl(5 + \sqrt3\bigr).$$

The problem asks for $$(PQ)^2,$$ so we square this expression:

$$$ \begin{aligned} (PQ_{\text{max}})^2 &= \left[\dfrac{1}{\sqrt2}\bigl(5 + \sqrt3\bigr)\right]^2 \\ &= \dfrac{1}{2}\bigl(5 + \sqrt3\bigr)^2 \\ &= \dfrac12\bigl(25 + 10\sqrt3 + 3\bigr) \\ &= \dfrac12\bigl(28 + 10\sqrt3\bigr) \\ &= 14 + 5\sqrt3. \end{aligned} $$$

Thus the maximum possible value of $$(PQ)^2$$ is $$14 + 5\sqrt3$$.

Hence, the correct answer is Option C.