Question

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?

• A. $CP$ is the arithmetic mean of $AP$ and $BP$.
• B. $PR$ is the geometric mean of $PS$ and $PC$.
• C. $PS$ is the harmonic mean of $PA$ and $PB$.
• D. The angle between the two tangents from $P$ is ${\tan }^{-1}\left(\frac{3}{4}\right)$.

Answer 1

Answer: (A), (B), (C)

$PS$ is the harmonic mean of $PA$ and $PB$.

Radius $=\sqrt{16+81-93}=2$
$CP=\sqrt{20}$
$\therefore AP=\sqrt{20}-2;BP=\sqrt{20}+2$
$\Rightarrow CP=\frac{AP+BP}{2}$


Now, let $L=PR=\sqrt{(PC{)}^2-r^2}=4=PQ$
$\tan \theta =\frac{2}{4}=\frac{1}{2}$
Also, $\cos \theta =\frac{PS}{PR}$
$\Rightarrow PS=PR\cos \theta =4\cdot \frac{2}{\sqrt{5}}=\frac{8}{\sqrt{5}}$

Harmonic mean between $PA$ and $PB$ is
$\frac{2(\sqrt{20}-2)(\sqrt{20}+2)}{2\sqrt{20}}=\frac{16}{2\sqrt{5}}=\frac{8}{\sqrt{5}}=PS$
Hence, $\left(PS\right)\left(PC\right)=16=(PR{)}^2$
$\Rightarrow PR$ is the geometric mean of $PS$ and $PC$

Now, the angle between the two tangents is $2\theta$ and $\tan \theta =\frac{1}{2}$
$\therefore 2{\tan }^{-1}\left(\frac{1}{2}\right)={\tan }^{-1}\frac{4}{3}$