If a point P has coordinates (0, -2) and Q is | vedclass

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(B) The equation of the circle is $x^2 + y^2 - 5x - y + 5 = 0$.Completing the square,we get $(x - 5/2)^2 + (y - 1/2)^2 = 25/4 + 1/4 - 5 = 6/4 = 3/2$.T

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$14 + 5\sqrt{3}$

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(B) The equation of the circle is $x^2 + y^2 - 5x - y + 5 = 0$.Completing the square,we get $(x - 5/2)^2 + (y - 1/2)^2 = 25/4 + 1/4 - 5 = 6/4 = 3/2$.The center of the circle is $C = (5/2, 1/2)$ and the radius is $r = \sqrt{3/2}$.The distance $PQ$ is maximized when $Q$ lies on the line passing through $P$ and $C$,specifically at the point on the circle furthest from $P$.The distance $PC$ is $\sqrt{(5/2 - 0)^2 + (1/2 - (-2))^2} = \sqrt{25/4 + 25/4} = \sqrt{50/4} = \frac{5\sqrt{2}}{2}$.The maximum distance $PQ$ is $PC + r = \frac{5\sqrt{2}}{2} + \sqrt{\frac{3}{2}} = \frac{5\sqrt{2} + \sqrt{6}}{2}$.Therefore,the maximum value of $(PQ)^2$ is $\left( \frac{5\sqrt{2} + \sqrt{6}}{2} \right)^2 = \frac{50 + 6 + 10\sqrt{12}}{4} = \frac{56 + 20\sqrt{3}}{4} = 14 + 5\sqrt{3}$.

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

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