Answer
$y=1$, see figure.
Work Step by Step
Given the curve as $y=(x-1)^2+1$ and a point $(1,1)$ on the curve, we can find the slope of the tangent line at this point as:
$m=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}=\lim_{h\to0}\frac{(1+h-1)^2+1-((1-1)^2+1))}{h}=\lim_{h\to0}\frac{h^2}{h}=\lim_{h\to0}(h)=0$
The equation for the tangent line at $(1,1)$ can then be written as $y-1=0(x-1)$ or $y=1$, as shown in the figure.