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

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$8 + 5 \sqrt{3}$

$\frac{47 + 10 \sqrt{6}}{2}$

$14 + 5 \sqrt{3}$

$\frac{25 + \sqrt{6}}{2}$

answer is C.

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Detailed Solution

Given circle is $x^{2} + y^{2} - 5 x - y + 5 = 0 \Rightarrow Centre = (\frac{5}{2}, \frac{1}{2}), radius = \sqrt{\frac{15}{4} + \frac{1}{4} - 5} = \sqrt{\frac{3}{2}}$

parametric form of circle is $x = \frac{5}{2} + \sqrt{\frac{3}{2}} \cos θ, y = \frac{1}{2} + \sqrt{\frac{3}{2}} \sin θ$

$\begin{array}{l} ∴ Q = (\frac{5}{2} + \sqrt{\frac{3}{2}} \cos θ, \frac{1}{2} + \sqrt{\frac{3}{2}} \sin θ) \\ P Q^{2} = {(\frac{5}{2} + \sqrt{\frac{3}{2}} \cos θ)}^{2} + {(\frac{5}{2} + \sqrt{\frac{3}{2}} \sin θ)}^{2} \\ = \frac{25}{4} + \frac{3}{2} \cos^{2} θ + 5 \sqrt{\frac{3}{2}} \cos θ + \frac{25}{4} + \frac{3}{2} \sin^{2} θ + 5 \sqrt{\frac{3}{2}} \sin θ \\ = \frac{25}{2} + \frac{3}{2} + 5 \frac{\sqrt{3}}{2} (\cos θ + \sin θ) . Maximum value of P Q^{2} is 14 + 5 \sqrt{\frac{3}{2}} (\sqrt{2}) = 14 + 5 \sqrt{3} . \end{array}$