Chebyshev expansion of $f(x)=\frac{1}{1+(x-s)^2}$

Question

The Chebyshev polynomials of the first kind are obtained from the recurrence relation $$\begin{aligned}T_{0}(x)&=1\\T_{1}(x)&=x\\T_{n+1}(x)&=2x\,T_{n}(x)-T_{n-1}(x)~.\end{aligned}$$ I would like to find the Chebyshev expansion (of the first kind), in $x$, of the function $$f(x)=\frac{1}{1+(x-s)^2}$$ where $x,s\in[-1,1]$ and $s$ is a parameter.