1. Write the circle in standard form

Given circle: $x^2+y^2-5x-y+5=0$

Complete the squares: $x^2-5x+y^2-y+5=0$ $\left(x-\frac{5}{2}\right)^2-\frac{25}{4}+\left(y-\frac{1}{2}\right)^2-\frac{1}{4}+5=0$ $\left(x-\frac{5}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=\frac{25}{4}+\frac{1}{4}-5=\frac{26}{4}-\frac{20}{4}=\frac{6}{4}=\frac{3}{2}$

So the circle has:

• Centre $C\left(\frac{5}{2},\frac{1}{2}\right)$
• Radius $r=\sqrt{\frac{3}{2}}=\frac{\sqrt6}{2}$

2. Find distance from $P(0,-2)$ to the centre

$PC^2=\left(\frac{5}{2}-0\right)^2+\left(\frac{1}{2}-(-2)\right)^2$ $=\left(\frac{5}{2}\right)^2+\left(\frac{5}{2}\right)^2=\frac{25}{4}+\frac{25}{4}=\frac{25}{2}$

Hence, $PC=\sqrt{\frac{25}{2}}=\frac{5}{\sqrt2}$

3. Maximum distance from an external point to a point on the circle

For any point $Q$ on the circle, the maximum value of $PQ$ occurs when $Q$ lies on the line joining $P$ and the centre, on the farthest side of the circle.

Thus, $PQ_{\max}=PC+r$

So, $PQ_{\max}=\frac{5}{\sqrt2}+\frac{\sqrt6}{2}$

Now we need $\left(PQ\right)^2_{\max}$: $\left(\frac{5}{\sqrt2}+\frac{\sqrt6}{2}\right)^2$

Write $\frac{5}{\sqrt2}=\frac{5\sqrt2}{2}$, so $PQ_{\max}=\frac{5\sqrt2+\sqrt6}{2}$

Then, $\left(PQ\right)^2_{\max}=\frac{(5\sqrt2+\sqrt6)^2}{4}$ $=\frac{50+6+2\cdot 5\sqrt2\cdot \sqrt6}{4}$ $=\frac{56+10\sqrt{12}}{4}$ $=\frac{56+20\sqrt3}{4}$ $=14+5\sqrt3$

4. Check options

The maximum value of $(PQ)^2 = 14+5\sqrt3$

So the correct option is: B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.