JEE Main 2021 — Circle Question with Solution

Let, the angle between the tangents $PA$ and $PB$ from the point $P$ to the circle $x^2+y^2-2x-4y+4=0$ with centre $C$ is $\theta .$

Given $\theta ={\tan }^{-1}\left(\frac{12}{5}\right) \Rightarrow \tan \theta =\frac{12}{5}.$

We know that the centre and radius of a circle $x^2+y^2+2gx+2fy+c=0$ are respectively $\left(-g, -f\right)$ and $\sqrt{g^2+f^2-c}.$

Thus, the centre and radius of the circle $x^2+y^2-2x-4y+4=0$ are respectively $\left(1, 2\right)$ and $\sqrt{1^2+2^2-4}=1$ unit.

We know that the line joining the centre of a circle to the point of contact of the tangent is perpendicular to the tangent, hence $\angle CAP=\angle CBP=90°$ also $CP$ bisects the angle $\angle APB$ as $∆CAP\cong ∆CBP$ by $SSS$ congruency.

Thus, we have $\angle CPA=\angle CPB=\frac{\theta }{2}$

In $∆CAP,$ we have $\tan \left(\frac{\theta }{2}\right)=\frac{AC}{AP}$

$\Rightarrow \tan \left(\frac{\theta }{2}\right)=\frac{1}{AP}$

$\Rightarrow AP=cot\left(\frac{\theta }{2}\right)$

Now, area of $\Delta PAB=\frac{1}{2}\left(PA\right)\left(PB\right)\sin \theta$

Since, tangents to a circle from an external point are equal, hence $PA=PB$

Area of $\Delta PAB=\frac{1}{2}{\left(PA\right)}^2\sin \theta$

$=\frac{1}{2}co{t}^2\left(\frac{\theta }{2}\right)\sin \theta$

$=\frac{1}{2}\frac{{\cos }^2\left({\displaystyle \frac{\theta }{2}}\right)}{{\sin }^2\left({\displaystyle \frac{\theta }{2}}\right)}\sin \theta$

Using $\cos 2x=2{\cos }^{2}x-1=1-2{\sin }^{2}x,$ we get

Area of $\Delta PAB=\frac{1}{2}\left(\frac{1+\cos \theta }{1-\cos \theta }\right)\sin \theta$

We have $\tan \theta =\frac{12}{5} \Rightarrow \sin \theta =\frac{12}{13} \& \cos \theta =\frac{5}{13}$

Area of $\Delta PAB=\frac{1}{2}\left(\frac{1+{\displaystyle \frac{5}{13}}}{1-{\displaystyle \frac{5}{13}}}\right)\left(\frac{12}{13}\right)$

$=\frac{1}{2}\times \frac{18}{8}\times \frac{12}{13}=\frac{27}{26}$ sq units.

And, area of $\Delta CAB=\frac{1}{2}\left(AC\right)\left(BC\right)\sin \left(\mathrm{\pi }-\mathrm{\theta }\right)$

$=\frac{1}{2}\sin \theta =\frac{1}{2}\left(\frac{12}{13}\right)=\frac{6}{13}$ sq units.

$\mathrm{\therefore } \frac{ area of \Delta PAB}{ area of \Delta CAB}=\frac{\left({\displaystyle \frac{27}{26}}\right)}{\left({\displaystyle \frac{6}{13}}\right)}=\frac{9}{4}.$