Tangents drawn from the point ( 4, 3) to the circle  $x^2+y^2-2x-4y=0$ are inclined at an angle 

using $S{S}_1=T^2$ the combined equation of the tangents drawn from (4, 3) to the circle $x^2+y^2-2x-4y=0$ is

$\begin{array}{r}\left(x^2+y^2-2x-4y\right)(16+9-8-12)\\ =\{4x+3y-(x+4)-2(y+3){\}}^2\end{array}$

$\Rightarrow 5\left(x^2+y^2-2x-4y\right)=(3x+y-10{)}^2$

$\Rightarrow 4x^2-4y^2-50x+6xy+100=0$          …(i)

In this equation, we have

$\text{ Coeff. of }{x}^2+\text{ Coeff. of }{y}^2=0$

Therefore, lines given by (i) are at right angle to each other. 

$\underline{ALITER}$ The equation of the given circle and its director circle

are

$(x-1{)}^2+(y-2{)}^2=5$ and $(x-1{)}^2+(y-2{)}^2=10$           …(i)

Clearly, $(4, 3)$ lies on (i). So, required angle is $\mathrm{\pi }/2$