$\text{ Consider the circle }{x}^2+y^2-8x-18y+93=0\text{ with centre ' }{C}^{\text{'}}\text{ and point }P(2,5)$outside it. From the point P, a pair of tangents PQ and PR are drawn to the circle with S as the midpoint of QR. The line joining P to C intersects the given circle at A and B. Which of the following hold(s) good?

$\begin{array}{l}\text{ Radius }=\sqrt{16+81-93}=2\\ \mathrm{CP}=\sqrt{20};\mathrm{AP}=\sqrt{20}-2;\mathrm{BP}=\sqrt{20}+2\\ \Rightarrow \mathrm{CP}=\frac{\mathrm{AP}+\mathrm{BP}}{2}\Rightarrow \mathrm{A}\text{ is correct }\end{array}$

$\begin{array}{l}\text{ Now, let }L=PR=\sqrt{(PC{)}^2-r^2}=\sqrt{20-4}=4=PQ;\tan \theta =\frac{2}{4}=\frac{1}{2}\\ \text{ Also }\cos \theta =\frac{\mathrm{PS}}{\mathrm{PR}}\Rightarrow \mathrm{PS}=\mathrm{PRcos}\theta =4.\left(\frac{2}{\sqrt{5}}\right)=\frac{8}{\sqrt{5}}\\ \text{ Harmonic Mean between PA and PB }=\frac{2(\sqrt{20}-2)(\sqrt{20}+2)}{2\sqrt{20}}=\frac{16}{2\sqrt{5}}=\frac{8}{\sqrt{5}}=\mathrm{PS}\Rightarrow \mathrm{C}\text{ is correct}\end{array}$

$\begin{array}{l}\text{ Hence }\left(\mathrm{PS}\right)\left(\mathrm{PC}\right)=\left(\frac{8}{\sqrt{5}}\right)\left(\sqrt{20}\right)=16=(\mathrm{PR}{)}^2\\ \Rightarrow \text{ PR is the Geometric Mean of }PS\text{ and }PC\Rightarrow B\text{ is correct }\end{array}$

$\begin{array}{l}\text{ Now angle between the two tangents is }2\theta \text{, then }\left(\text{ As }\tan \theta =\frac{1}{2}\right)\\ 2{\tan }^{-1}\left(\frac{1}{2}\right)={\tan }^{-1}\left(\frac{2\cdot \frac{1}{2}}{1-\frac{1}{4}}\right)={\tan }^{-1}\left(\frac{4}{3}\right)\Rightarrow \mathrm{D}\text{ is incorrect }\end{array}$