Extremum Problem 

How big is the circle?   

My first steps: 


*

*$ x^2 + y^2 =r^2$   

*$ f(x)=y=e^{-x^2}$    


Substitute $y^2$ in $x^2 + y^2 =r^2.$
So, $x^2 + e^{-2x^2} =r^2$  
Is this way correct? Because after calculating the first derivative $2x -4xe^{-2x^2}$  and so on my solution at the end is $x^2 = \dfrac {\ln(2)}{2}, $ and this is not correct.

Your solution is right. Clearly the circle and the curve meet at $(x,y)$ in first quadrant where $$x^2+y^2=r^2\\y=e^{-x^2}\\-2xe^{-x^2}=\dfrac{-x}{\sqrt{r^2-x^2}}$$which means that $$2y=\dfrac{1}{\sqrt {r^2-x^2}}=\dfrac{1}{y}$$therefore $$y=\dfrac{\sqrt 2}{2}$$and $$x=\sqrt{\dfrac{\ln 2}{2}}$$which means that $$r=\sqrt{\dfrac{1+\ln 2}{2}}\approx 0.9200943377$$
