Question

$C_1\text{and}{C}_2$ are circles of unit radius with centres at $(0,0)\text{and}(1,0)$ respectively. $C_3$ is a circle of radius $r(r\in N)$. It passes through the centres of the circles $C_1\text{and}{C}_2$ and has its center above the $x-$axis. If the common tangent of $C_1\text{and}{C}_3$ is $\sqrt{3}x-y+c=0$. Then maximum value of $c+r=$

• A. $3$
• B. $4$
• C. $2$
• D. $1$

Answer 1

The correct option is A $3$

Equation of the $C_3:(x-a{)}^2+(y-b{)}^2=r^2$
$\because$ it passes through $(0,0)\text{and}(1,0)\therefore a^2+b^2=r^2...\left(1\right)(a-1{)}^2+b^2=r^2...(2)\Rightarrow a=\frac{1}{2},b=\pm \frac{\sqrt{4r^2-1}}{2}\to b=\frac{\sqrt{4r^2-1}}{2}{C}_3:x^2+y^2-x-(\sqrt{4r^2-1})y=0$
$\because$ equation of the common tangents is $\sqrt{3}x-y+c=0$
Distance of the tangent from the origin is $1$
$\therefore \frac{|c|}{\sqrt{3+1}}=1\Rightarrow c=\pm 2$
$\text{for}c=2$
$\frac{\sqrt{3}a-b+2}{2}=r\Rightarrow \sqrt{3}+4-\sqrt{4r^2-1}=4r\Rightarrow 2\sqrt{3}r+8r=3r^2+5+2\sqrt{3}\Rightarrow (r-1)(3r-5-2\sqrt{3})=0\therefore r=1$
$\text{for}c=-2$
$\frac{-(\sqrt{3}a-b-2)}{2}=r\Rightarrow \sqrt{3}-4-\sqrt{4r^2-1}=-4r\Rightarrow 12r^2+20-32r+8\sqrt{3}r-8\sqrt{3}=0\Rightarrow (r-1)(3r-5-2\sqrt{3})=0\therefore r=1$