Test Index

Circles Part 1

Section: Mathematics

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Question : 19 of 100

Marks: +1, -0

Two tangents are drawn from a point

P

x^{2}+y^{2}-2 x-4 y+4=0

, such that the angle between these tangents is

\tan^{-1}\left(\;\frac{12}{5}\right)

\tan^{-1}\left(\;\frac{12}{5}\right) \in (0, \pi)

. If the centre of the circle is denoted by

C

and these tangents touch the circle at points

A

B

, then the ratio of the areas of

\triangle PAB

\triangle CAB

[17 Mar 2021 Shift 2]

11: 4
$2:1$
$3: 1$
$2: 1$

Solution:

Method (I)

Given equation of cricle

x^{2}+y^{2}-2 x-4 y+4=0

Centre

(1,2), R=\sqrt{1+4-4}=1

Let

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

\;\frac{2 \tan\left(\frac{\theta}{2}\right)}{1-\tan^2\left(\;\frac{\theta}{2}\right)}=\;\frac{12}{5}

\Rightarrow 6 \tan^2\left(\;\frac{\theta}{2}\right)+5 \tan\left(\;\frac{\theta}{2}\right)-6=0

\Rightarrow \;\; \left(3 \tan \;\frac{\theta}{2}-2\right)\left(2 \tan \;\frac{\theta}{2}+3\right)=0

Either

\tan \;\frac{\theta}{2}=\;\frac{2}{3}

or

\tan \;\frac{\theta}{2}=\;\frac{-3}{2}

(Rejected) ...(i)

\left[ \text{ }\; \theta \in (0, \pi) , \therefore \;\frac{\theta}{2} \in \left(0, \;\frac{\pi}{2}\right) , \therefore \tan \;\frac{\theta}{2}>0 \right]

From figure,

CA=CB=R=1

\text{In } \triangle CAP, \tan \;\frac{\theta}{2}=\;\frac{CA}{AP}=\;\frac{1}{L}

...(ii)

From Eqs. (i) and (ii),

\;\frac{1}{L}=\;\frac{2}{3} \Rightarrow L=\;\frac{3}{2}

Let

\;\; \angle ACP=\phi

\therefore \;\; \phi=\;\frac{\pi}{2}-\theta \Rightarrow \tan \phi=\tan\left(\;\frac{\pi}{2}-\;\frac{\theta}{2}\right)

\Rightarrow \;\; \tan \phi=\cot \;\frac{\theta}{2} \Rightarrow \tan \phi=\;\frac{3}{2}

[from Eq. (i)]

\therefore \;\; \cos \phi=\;\frac{2}{\sqrt{13}}

and

\sin \phi=\;\frac{3}{\sqrt{13}}

In

\triangle AMC

,

\sin \phi=\;\frac{P_{1}}{1}

and

\cos \phi=\;\frac{CM}{1} \Rightarrow P_{1}=\;\frac{3}{\sqrt{13}} \Rightarrow CM=\;\frac{2}{\sqrt{13}}

Now, area of

\triangle CAM=\;\frac{1}{2} \times \;\frac{2}{\sqrt{13}} \times \;\frac{3}{\sqrt{13}}=\;\frac{3}{13}

\therefore

Area of

\triangle CAB=2(\triangle CAM)=\;\frac{6}{13}

Area of

\triangle PACBP=2 \times (\triangle PAC)

Now, area of

=2 \times \;\frac{1}{2} \times L \times R=RL=\;\frac{3}{2}

Area of

\therefore

\triangle PAB=RL-

\triangle CAB

Area of

=\;\frac{3}{2}-\;\frac{6}{13}=\;\frac{39-12}{26}=\;\frac{27}{26}

Area of

\;\frac{ \;\text{ Area of }\; \triangle PAB }{ \;\text{ Area of }\; \triangle CAB }=\;\frac{ \frac{27}{26} }{ \frac{6}{13} }=\;\frac{9}{4}

9:4

Hence,

3:1

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