Question

$3$ circle of radii $1,2$ and $3$ and centres at $A,B$ and $C$ respectively, touch each other. Another circle whose centre is $P$ touches all these 3 circles externally and has radius $r$. Also $\angle PAB=\theta$ & $\angle PAC=\alpha$ -

• A. $\cos \theta =\frac{3-r}{3(1+r)}$
• B. $\cos \alpha =\frac{2-r}{2(1+r)}$
• C. $r=\frac{6}{23}$
• D. $r=\frac{6}{\sqrt{23}}$

Answer 1

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

$△ABC$ is right angle
Applying cosine rule in $△PAB$
$\cos \theta =\frac{3^2+(1+r{)}^2-(2+r{)}^2}{2\cdot 3(1+r)}$
$=\frac{3-r}{3(1+r)}$
Again applying cosine rule in $△PAC$
$\cos \alpha =\frac{(1+r{)}^2+4^2-(3+r{)}^2}{2\cdot 4(1+r)}=\frac{2-r}{2(1+r)}$
$\because \alpha +\theta =90$
$\alpha =90-\theta \Rightarrow \cos \alpha =\sin \theta$
${\left(\frac{3-r}{3(r+1)}\right)}^2+{\left(\frac{2-r}{2(r+1)}\right)}^2=1$