Let ABCD be a square of side of unit length. Let a circle C _ 1...

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Here, $AO + OP = 1$ or \left(\sqrt{2} + 1\right)r = 1 ⇒ r = \dfrac{1}{\sqrt{2} + 1} \times \dfrac{\sqrt{2} - 1}{\sqrt{2} - 1} = \sqrt{2} - 1$.</p><p>The equation of circle is${\left(x - r\right)}^{2} + {\left(y - r\right)}^{2} = {r}^{2}$. </p><p>The equation of tangent$CE$of slope$m$is$y - 1 = m\left(x - 1\right)$</p><p>$⇒ mx - y + 1 - m = 0$</p><p>It is tangent to the circle, then the distance of center of the circle${C}_{2}\left(r, r\right)$from the tangent$CE$is equal to the radius of circle$r$.</p><p>$∴ \left|\dfrac{mr - r + 1 - m}{\sqrt{{m}^{2} + 1}}\right| = r$</p><p>$⇒ \left|\dfrac{\left(m - 1\right)r + 1 - m}{\sqrt{{m}^{2} + 1}}\right| = r$</p><p>$⇒ \dfrac{{\left(m - 1\right)}^{2}{\left(r - 1\right)}^{2}}{{m}^{2} + 1} = {r}^{2}$</p><p>Put$r = \sqrt{2} - 1$</p><p>On solving, we get</p><p>$m = 2 - \sqrt{3}, 2 + \sqrt{3}$</p><p>Taking greater slope of$CE$as$2 + \sqrt{3}$, we get </p><p>Equation of tangent as$y - 1 = \left(2 + \sqrt{3}\right)\left(x - 1\right)$</p><p>By substituting$y = 0$, we get$x$-coordinate of$E$,</p><p>$- 1 = \left(2 + \sqrt{3}\right)\left(x - 1\right)$</p><p>$\dfrac{- 1}{2 + \sqrt{3}} \times \left(\dfrac{2 - \sqrt{3}}{2 - \sqrt{3}}\right) = x - 1$</p><p>$x = \sqrt{3} - 1$</p><p>So the coordinates of$E\text{are}\left(\sqrt{3}, 0\right)$</p><p>$∴ EB = 1 - x = 1 - \left(\sqrt{3} - 1\right)$</p><p>$⇒ EB = 2 - \sqrt{3}