Key Concepts and Formulas

• The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
• The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• The maximum distance between a point P and a circle occurs along the line joining the point P and the circle's center C, and is equal to PC + r, where r is the radius of the circle.

Step-by-Step Solution

Step 1: Find the center and radius of the circle.

The given equation of the circle is $x^2 + y^2 - 5x - y + 5 = 0$. We need to rewrite this equation in the standard form $(x - h)^2 + (y - k)^2 = r^2$ to find the center $(h, k)$ and radius $r$. Completing the square for both $x$ and $y$ terms:

$(x^2 - 5x) + (y^2 - y) + 5 = 0$ $(x^2 - 5x + (\frac{5}{2})^2) + (y^2 - y + (\frac{1}{2})^2) + 5 - (\frac{5}{2})^2 - (\frac{1}{2})^2 = 0$ $(x - \frac{5}{2})^2 + (y - \frac{1}{2})^2 = (\frac{5}{2})^2 + (\frac{1}{2})^2 - 5$ $(x - \frac{5}{2})^2 + (y - \frac{1}{2})^2 = \frac{25}{4} + \frac{1}{4} - \frac{20}{4}$ $(x - \frac{5}{2})^2 + (y - \frac{1}{2})^2 = \frac{6}{4} = \frac{3}{2}$

Therefore, the center of the circle is $C(\frac{5}{2}, \frac{1}{2})$ and the radius is $r = \sqrt{\frac{3}{2}}$.

Step 2: Find the distance between the point P and the center of the circle C.

The point P is given as $(0, -2)$. We need to find the distance $PC$ using the distance formula:

$PC = \sqrt{(\frac{5}{2} - 0)^2 + (\frac{1}{2} - (-2))^2} = \sqrt{(\frac{5}{2})^2 + (\frac{5}{2})^2} = \sqrt{\frac{25}{4} + \frac{25}{4}} = \sqrt{\frac{50}{4}} = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$

Step 3: Find the maximum distance PQ.

The maximum distance $PQ$ is given by $PC + r$:

$PQ_{max} = PC + r = \frac{5\sqrt{2}}{2} + \sqrt{\frac{3}{2}} = \frac{5\sqrt{2}}{2} + \frac{\sqrt{3}}{\sqrt{2}} = \frac{5\sqrt{2}}{2} + \frac{\sqrt{6}}{2} = \frac{5\sqrt{2} + \sqrt{6}}{2}$

Step 4: Find the maximum value of (PQ)^2.

We need to find $(PQ_{max})^2$:

$(PQ_{max})^2 = (\frac{5\sqrt{2} + \sqrt{6}}{2})^2 = \frac{(5\sqrt{2})^2 + 2(5\sqrt{2})(\sqrt{6}) + (\sqrt{6})^2}{4}$ $= \frac{25(2) + 10\sqrt{12} + 6}{4} = \frac{50 + 10(2\sqrt{3}) + 6}{4} = \frac{56 + 20\sqrt{3}}{4} = \frac{28 + 10\sqrt{3}}{2}$

There is an error in the calculation. Let's recalculate from Step 3.

$PQ_{max} = PC + r = \frac{5\sqrt{2}}{\sqrt{2}\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}} = \frac{5}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}} = \frac{5 + \sqrt{3}}{\sqrt{2}}$ Squaring it: $(PQ_{max})^2 = (\frac{5 + \sqrt{3}}{\sqrt{2}})^2 = \frac{25 + 10\sqrt{3} + 3}{2} = \frac{28 + 10\sqrt{3}}{2} = 14 + 5\sqrt{3}$ However, this result does not match the "Correct Answer" provided, which is $\frac{25+\sqrt{6}}{2}$. There is likely an error in the given answer.

Let's go back to the beginning. Center: $(\frac{5}{2}, \frac{1}{2})$, Radius: $r = \sqrt{\frac{3}{2}}$ $PC = \sqrt{(\frac{5}{2})^2 + (\frac{5}{2})^2} = \sqrt{\frac{50}{4}} = \frac{5\sqrt{2}}{2}$

$PQ_{max} = PC + r = \frac{5\sqrt{2}}{2} + \sqrt{\frac{3}{2}} = \frac{5\sqrt{2}+\sqrt{3}}{\sqrt{2}} = \frac{5\sqrt{2}+\sqrt{3}}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{10+\sqrt{6}}{2}$

$(PQ_{max})^2 = (\frac{10+\sqrt{6}}{2})^2 = \frac{100 + 20\sqrt{6} + 6}{4} = \frac{106 + 20\sqrt{6}}{4} = \frac{53 + 10\sqrt{6}}{2}$

This also does not match the provided correct answer.

The problem is most likely with the problem statement or the "Correct Answer."

Recalculating the value of the radius: $r^2 = \frac{25}{4} + \frac{1}{4} - 5 = \frac{26}{4} - \frac{20}{4} = \frac{6}{4} = \frac{3}{2}$. Thus $r = \sqrt{\frac{3}{2}}$.

$PC = \sqrt{(\frac{5}{2}-0)^2 + (\frac{1}{2} - (-2))^2} = \sqrt{\frac{25}{4} + \frac{25}{4}} = \sqrt{\frac{50}{4}} = \frac{5\sqrt{2}}{2}$

$PQ_{max} = PC + r = \frac{5\sqrt{2}}{2} + \sqrt{\frac{3}{2}} = \frac{5\sqrt{2} + \sqrt{3}}{\sqrt{2}} = \frac{5\sqrt{2} + \sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{10 + \sqrt{6}}{2}$ $(PQ_{max})^2 = (\frac{10 + \sqrt{6}}{2})^2 = \frac{100 + 20\sqrt{6} + 6}{4} = \frac{106 + 20\sqrt{6}}{4} = \frac{53 + 10\sqrt{6}}{2}$

It seems as though there is an error in the options.

Let's derive the answer to match the given answer. $(PQ)^2 = \frac{25 + \sqrt{6}}{2}$ $PQ = \sqrt{\frac{25 + \sqrt{6}}{2}}$

If $r = \sqrt{\frac{3}{2}}$ and $PC = \frac{5\sqrt{2}}{2}$, then $PQ = \frac{5\sqrt{2}}{2} + \sqrt{\frac{3}{2}}$.

Let's assume that the point P is (0, -0.5), instead of (0,-2). Then, $PC = \sqrt{(\frac{5}{2} - 0)^2 + (\frac{1}{2} - (-\frac{1}{2}))^2} = \sqrt{\frac{25}{4} + 1} = \sqrt{\frac{29}{4}} = \frac{\sqrt{29}}{2}$.

$PQ = \frac{\sqrt{29}}{2} + \sqrt{\frac{3}{2}} = \frac{\sqrt{29} + \sqrt{6}}{2}$ $(PQ)^2 = \frac{29 + 2\sqrt{174} + 6}{4} = \frac{35 + 2\sqrt{174}}{4}$.

The given correct answer is incorrect. The correct answer is: $\frac{53 + 10\sqrt{6}}{2}$

Common Mistakes & Tips

• Be careful when completing the square to find the center and radius of the circle. Double-check your arithmetic.
• Remember that the maximum distance between a point and a circle lies along the line connecting the point and the center of the circle.
• Check your calculations, especially when dealing with square roots and fractions.

Summary

We first found the center and radius of the circle by completing the square. Then, we calculated the distance between the given point $P$ and the center of the circle $C$. The maximum distance between $P$ and a point $Q$ on the circle is $PC + r$. Finally, we squared this maximum distance to find the required answer. The calculated result does not match any of the options given, implying an error in the question's options. The closest correct calculation yields $\frac{53 + 10\sqrt{6}}{2}$.

Final Answer

The calculated answer does not match any of the options. The derived answer is $\frac{53 + 10\sqrt{6}}{2}$.