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

Equation of tangent $=3x+4y=6$ and two normals are

$(x-1)(y-2)=0$

$\Rightarrow x-1=0\,$ and $\, y-2=0$

$\therefore \text{Radius}=\displaystyle \frac{3(1)+4(2)-6}{\sqrt{9+16}}$

$=\displaystyle \frac{5}{5}=1$

$\therefore$ Equation of the circle is

$(x-1)^2+(y-2)^2=1\,$ or$\, x^2+y^2-2x-4y+4=0$