Question

Let $ABCD$ be a square of side of unit length. Let a circle $C_1$ centered at $A$ with unit radius is drawn. Another circle $C_2$ which touches $C_1$ and the lines $AD$ and $AB$ are tangent to it, is also drawn. Let a tangent line from the point $C$ to the circle $C_2$ meet the side $AB$ at $E$. If the length of $EB$ is $\alpha +\sqrt{3}\beta$, where $\alpha ,\beta$ are integers, then $\alpha +\beta$ is equal to

Answer 1

Answer: 1

Here $AO+OD=1$ or $(\sqrt{2}+1)r=1$
$\Rightarrow r=\sqrt{2-1}$
equation of circle $(x-r{)}^2+(y-r{)}^2=r^2$
Equation of $CE$
$y-1=m(x-1)$
$mx-y+1-M=0$
It is tangent to circle
$\therefore \left|\frac{mr-r+1-m}{\sqrt{m^2+1}}\right|=r$
$\left|\frac{(m-1)r+1-m}{\sqrt{m^2+1}}\right|=r$
$\frac{(m-1{)}^2(r-1{)}^2}{m^2+1}=r^2$
Put $r=\sqrt{2}-1$
On solving $m=2-\sqrt{3},2+\sqrt{3}$
Taking greater slope of $CE$ as $2+\sqrt{3}$
$y-1=(2+\sqrt{3})(x-1)$
Put $y=0$
$-1=(2+\sqrt{3})(x-1)$
$\frac{-1}{2+\sqrt{3}}\times \left(\frac{2-\sqrt{3}}{2-\sqrt{3}}\right)=x-1$
$x-1=\sqrt{3}-1$
$EB=1-x=1-(\sqrt{3}-1)$
$EB=2-\sqrt{3}$