The point of intersection of the common tangents drawn to the circles $x^2+y^2-4x-2y+1=0\text{ and }{x}^2+y^2-6x-4y+4=0$.

For the circle $x^2+y^2-4x-2y+1=0;\text{ centre }{C}_1=(2,1),\text{ radius }{r}_1=2$

For the circle $x^2+y^2-6x-4y+4=0;\text{ centre }{C}_2=(3,2),\text{ radius }{r}_2=3$

$\therefore$ Point of intersection of common tangents : external centre of similitude

=The point which divides $\overline{C_1{C}_2}\text{ in the ratio }-2:3=\left(\frac{-6+6}{-2+3},\frac{-4+3}{-2+3}\right)=(0,-1)\text{. }$