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 : $r=\sqrt{16+81-93}=2$
$CP=\sqrt{20};AP=\sqrt{20}-2;BP=\sqrt{20}+2$
$\frac{AP+BP}{2}=\sqrt{20}=CP$
Hence, $CP$ is the arithmetic mean of $AP$ and $BP$.

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

Harmonic mean between $PA$ and $PB$
$=\frac{2(\sqrt{20}-2)(\sqrt{20}+2)}{\sqrt{20}-2+\sqrt{20}+2}=\frac{16}{2\sqrt{5}}=\frac{8}{\sqrt{5}}=PS$
Hence, $PS$ is the harmonic mean of $PA$ and $PB$.

Geometric mean of $PS$ and $PC$ $=\sqrt{\left(PS\right)\left(PC\right)}=\sqrt{\left(\frac{8}{\sqrt{5}}\right)\left(\sqrt{20}\right)}=4=PR$
Hence, $PR$ is the geometric mean of $PS$ and $PC$.

Angle between the two tangents is $2\theta .$
Then $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{2\cdot \frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3}$
$\Rightarrow 2\theta ={\tan }^{-1}\frac{4}{3}$