Let A B C D be a square of side of unit length. Let a circle C 1 centered at A with unit radius is drawn.…

According to the question we get,  

Here, $AO+OP=1$ or $\left(\sqrt{2}+1\right)r=1\Rightarrow r=\frac{1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}=\sqrt{2}-1$.

The equation of circle is ${\left(x-r\right)}^2+{\left(y-r\right)}^2=r^2$. 

The equation of tangent $CE$ of slope $m$is $y-1=m\left(x-1\right)$

$\Rightarrow mx-y+1-m=0$

It is tangent to the circle, then the distance of center of the circle $C_2\left(r,r\right)$ from the tangent $CE$ is equal to the radius of circle $r$.

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

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

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

Put $r=\sqrt{2}-1$

On solving, we get

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

Taking greater slope of $CE$ as $2+\sqrt{3}$, we get 

Equation of tangent as $y-1=\left(2+\sqrt{3}\right)\left(x-1\right)$

By substituting $y=0$, we get $x$-coordinate of $E$,

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

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

$x=\sqrt{3}-1$

So the coordinates of $E \text{are} \left(\sqrt{3},0\right)$

$\therefore EB=1-x=1-\left(\sqrt{3}-1\right)$

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