The equation of a circle has two normal $(x-1)(y-2)=0$  and a tangent $3x+4y=6$  is

$\begin{array}{l}(x-1)(x-2)=0\Rightarrow x-1=0\to \left(1\right)\\ y-2=0\to \left(2\right)\end{array}$

Point of intersection of (1) and (2) is $\begin{array}{l}(1,2)\\ hk\end{array}$ = centre

$r=$ the perpendicular distance from (1, 2) to the line $3x+4y=6$

$=\left|\frac{3\left(1\right)+4\left(2\right)-6}{\sqrt{3^2+4^2}}\right|=1$

The equation of a circle is ${(x-h)}^2+{(y-k)}^2=r^2$

$\begin{array}{l}{(x-1)}^2+{(y-2)}^2=1\\ \Rightarrow x^2+y^2-2x-4y+4=0\end{array}$