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

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 $CE$
$y -1= m ( x -1)$
$m x-y+1-M=0$
It is tangent to circle
$\therefore \left|\frac{m r-r+1-m}{\sqrt{m^{2}+1}}\right|=r$
$\left|\frac{(m-1) r+1-m}{\sqrt{m^{2}+1}}\right|=r$
$\frac{(m-1)^{2}(r-1)^{2}}{m^{2}+1}=r^{2}$
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 $y =0$
$-1=(2+\sqrt{3})( x -1)$
$\frac{-1}{2+\sqrt{3}} \times\left(\frac{2-\sqrt{3}}{2-\sqrt{3}}\right)= x -1$
$x-1=\sqrt{3}-1$
$EB =1- x =1-(\sqrt{3}-1)$
$EB =2-\sqrt{3}$