The total number of common tangents of $x^{2}+y^{2}-6 x-8 y+9=0$ and $x^{2}+y^{2}=1$ is

The given equation of circles and $$ \begin{gathered} x^{2}+y^{2}-6 x-8 y+9=0 \\ x^{2}+y^{2}=1 \end{gathered} $$ Radius and centre of both circles is $C_{1} \rightarrow(3,4), R_{1}=\sqrt{9+16-9}=4$ $C_{2} \rightarrow(0,0), R_{2}=1$ $C_{1} C_{2}=\sqrt{(3-0)^{2}+(4-0)^{2}}=\sqrt{9+16}=5$ $$ R_{1}+R_{2}=4+1=5 $$ $$ C_{1} C_{2}=R_{1}+R_{2} $$ distinct while the transverse tangents are coincident. So, number of comman tangents $=3$