# Let ABCD be a square of side of unit length.

Question:

Let $\mathrm{ABCD}$ be a square of side of unit length.

Let a circle $C_{1}$ centered at $A$ with unit radius is drawn. Another circle $\mathrm{C}_{2}$ which touches $\mathrm{C}_{1}$ and the lines $\mathrm{AD}$ and $\mathrm{AB}$ are tangent to it, is also drawn. Let a tangent line from the point $\mathrm{C}$ to the circle $\mathrm{C}_{2}$ meet the side $\mathrm{AB}$ at $\mathrm{E}$. If the length of $\mathrm{EB}$ is $\alpha+\sqrt{3} \beta$, where $\alpha, \beta$ areintegers, then $\alpha+\beta$ is equal to

Solution:

Here $\mathrm{AO}+\mathrm{OD}=1$ or $(\sqrt{2}+1) \mathrm{r}=1$

$\Rightarrow \quad \mathrm{r}=\sqrt{2-1}$

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

$\therefore \quad\left|\frac{\mathrm{mr}-\mathrm{r}+1-\mathrm{m}}{\sqrt{\mathrm{m}^{2}+1}}\right|=\mathrm{r}$

$\left|\frac{(m-1) r+1-m}{\sqrt{m^{2}+1}}\right|=r$

$\frac{(m-1)^{2}(r-1)^{2}}{m^{2}+1}=r^{2}$

Put $\mathrm{r}=\sqrt{2}-1$

On solving $\mathrm{m}=2-\sqrt{3}, 2+\sqrt{3}$

Taking greater slope of $C E$ as $2+\sqrt{3}$

$y-1=(2+\sqrt{3})(x-1)$

Put $\quad y=0$

$-1=(2+\sqrt{3})(\mathrm{x}-1)$

$\frac{-1}{2+\sqrt{3}} \times\left(\frac{2-\sqrt{3}}{2-\sqrt{3}}\right)=x-1$

$x-1=\sqrt{3}-1$

$\mathrm{EB}=1-\mathrm{x}=1-(\sqrt{3}-1)$

$\mathrm{EB}=2-\sqrt{3}$