Tangent graph to a circle

Solving the equation $f'(x)=g'(x)$ for $b$, you get $$ b = -x - \frac{\sqrt{1-x^2}}{x} $$ Plug that in to $f(x) = g(x)$ and you have $$ \ln \left(-\frac{\sqrt{1-x^2}}{x}\right) = \sqrt{1-x^2} $$ This is unlikely to have closed-form solutions, but you can use numerical methods, such as Newton's. Maple tells me $$ x = -0.3669416757$$ which corresponds to $$ b = 2.902069222$$