Consider the circle $x^2+y^2-8x-18y+93=0$ with centre ‘C’ 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?

We have, radius $=\sqrt{16+81-93}=2$

$$\begin{gathered} CP=\sqrt{20};AP=\sqrt{20}-2;BP=\sqrt{20}+2 \\ \Rightarrow CP=\frac{AP+BP}{2} \end{gathered}$$

Now, Let  $L=PR=\sqrt{{\left(PC\right)}^2-r^2}=\sqrt{20-4}=4=PQ$
and  $\tan \theta =\frac{2}{4}=\frac{1}{2}$
Also,  $\cos \theta =\frac{PS}{PR}$
$\Rightarrow PS=PR\cos \theta =4.\left(\frac{2}{\sqrt{5}}\right)=\frac{8}{\sqrt{5}}$

Harmonic mean between PA and PB

$$\begin{gathered} =\frac{2(\sqrt{20}-2)(\sqrt{20}+2)}{2\sqrt{20}}=\frac{16}{2\sqrt{5}}=\frac{8}{\sqrt{5}}=PS \\ Hence, \left(PS\right)\left(PC\right)=\left(\frac{8}{\sqrt{5}}\right)\left(\sqrt{20}\right)=16={\left(PR\right)}^2 \end{gathered}$$

$\Rightarrow PR$ is the Geometric Mean of PS and PC
Now angle between the two tangents is $2\theta$, then 

$$\begin{gathered} 2{\tan }^{-1}\left(\frac{1}{2}\right)={\tan }^{-1}\left(\frac{2.\frac{1}{2}}{1-\frac{1}{4}}\right)(As\tan \theta =\frac{1}{2}) \\ ={\tan }^{-1}\left(\frac{4}{3}\right) \end{gathered}$$