Let [math]CE=x,[/math] [math]AF=y[/math] and[br][math]OI=d[/math][br][math]CA[/math] is diameter[br][math]\Longrightarrow\angle CBA=90^\circ[/math] [br][math]AB[/math] and [math]BC[/math] are tangents[br][math]\Longrightarrow\angle CEI=\angle IFA=90^{\circ}[/math][br][math]\Longrightarrow[/math][math]IFBE[/math] is sqare, because [math]IF[/math] and [math]IE[/math] are radii[br][math]\Longrightarrow[/math][math]IF=FB=BE=EI=r[/math] and [math]\text{\triangle}CIE\sim\text{\triangle}IAF\sim\text{\triangle}CAB[/math][br][math]\Longrightarrow AB:IE:AF=BC:EC:FI=AC:IC:AI[/math][br] [math](y+r):r:y=(x+r):x:r=2:(1+d):(1-d)[/math][br]Pythagoras' theorem for [math]\text{\triangle}ABC[/math][br] [math]AC^2=AB^2+BC^2[/math][sup][br][/sup] [math]4=(r+y)^2+(x+r)^2[/math][sup][br][/sup]Pythagoras' theorem for[sup] [/sup][math]\text{\triangle}CIE[/math][sup][br] [/sup][math]CI^2=IE^2+CE^2[/math][sup][br] [/sup][math]\left(d+1\right)^2=r^2+x^2[/math][br]Pythagoras' theorem for [math]\text{\triangle}AIF[/math][br][sup] [/sup][math]AI^2=FI^2+FA^2[/math][sup][br] [/sup][math]\left(1-d\right)^2=r^2+y^2[/math][sup][br][/sup][math]\Longrightarrow[/math][math]4=r^2+2rx+x^2+r^2+2ry+y^2[/math][br] [math]4=(d+1)^2+(1-d)^2+2r(x+y)[/math][br] [math]4=d^2+2d+1+1-2d+d^2+2r(x+y)[/math][br] [math]1=d^2+r\left(x+y\right)[/math][br] [math]x+y=\frac{\left(1-d^2\right)}{r}[/math][br][math]\Longrightarrow[/math][math]\frac{y}{r}=\frac{1-d}{1+d}[/math] [math]y=r\frac{1-d}{1+d}[/math][br] [math]\frac{r}{y}=\frac{x}{r}[/math][math]\frac{x}{r}=\frac{1+d}{1-d}[/math][math]x=r.\frac{1+d}{1-d}[/math][br][math]\Longrightarrow x+y=r\text{ }\left(\frac{1+d}{1-d}+\frac{1-d}{1+d}\right)[/math][br][math]\Longrightarrow\frac{1-d^2}{r}=r\frac{2+2d^2}{\sqrt{2\left(1+d^2\right)}}[/math][br] [math]r=\frac{1-d^2}{\sqrt{2\left(1+d^2\right)}}[/math][br] [br][br]In conclusion it can be said that [math]r=\frac{1-d^2}{\sqrt{2\left(1+d^2\right)}}[/math].