The equation of the tangent at $(1,1)$ to circle $2x^2+2y^2-2x-5y+3=0$

$\begin{array}{l}{x}^2+y^2-x-\frac{5}{2}y+\frac{3}{2}=0\\ S_1=0,P(1,1)\\ x{x}_1+y{y}_1-(\frac{x}{2}+\frac{x_1}{2})-\frac{5}{4}(y+y_1)+\frac{3}{2}=0\\ x+y-(\frac{x}{2}+\frac{1}{2})-\frac{5}{4}(y+1)+\frac{3}{2}=0\end{array}$

$x+y-\frac{x}{2}-\frac{1}{2}-\frac{5}{4}y-\frac{5}{4}+\frac{3}{2}=0$ then it becomes

$2x-y-1=0$