A circle passes through A(1, 1) and touches x-axis then the locus of the other end of the diameter through A is 
 

$$\begin{gathered} Let the circle be {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 \\ It passes through A(1,1) then \\ 1+1+2\mathrm{g}+2\mathrm{f}+\mathrm{c}=0 \\ \Rightarrow 2+2g+2f+c=0\to \left(1\right) \\ and touches x-axis then {\mathrm{g}}^2=\mathrm{c} \\ \mathrm{A} (1, 1) ,\mathrm{other} \mathrm{end} be (\mathrm{x}, \mathrm{y}) \\ \Rightarrow \left(\frac{1+\mathrm{x}}{2}, \frac{1+\mathrm{y}}{2}\right)=(-\mathrm{g}, -\mathrm{f}) \\ 1+\mathrm{x}=-2\mathrm{g} \overline{) 1+\mathrm{y}=-2\mathrm{f}} \\ \mathrm{from} \left(1\right) \\ \Rightarrow 2-1-\mathrm{x}-1-\mathrm{y}+{\left(\frac{1+\mathrm{x}}{-2}\right)}^2=0 \\ \Rightarrow -\mathrm{x}-\mathrm{y}+\frac{1+{\mathrm{x}}^2+2\mathrm{x}}{4}=0 \\ \Rightarrow {\mathrm{x}}^2+2\mathrm{x}+1-4\mathrm{x}-4\mathrm{y}=0 \\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}=4\mathrm{y}-1 \\ \Rightarrow {(\mathrm{x}-1)}^2=4\mathrm{y} \end{gathered}$$