The number of common tangents to $x^2+y^2=256,{(x-3)}^2+{(y-4)}^2=121$ is

$$\begin{gathered} \mathrm{For} \mathrm{given} \mathrm{circles} \\ \mathrm{Centres} \mathrm{are} \\ {\mathrm{C}}_1=(0, 0) C{,}_2=(3, 4) \\ and radii are {\mathrm{r}}_1=16, {\mathrm{r}}_2=11 \\ {C}_1{C}_2=\sqrt{9+16}=5 \\ \therefore {\mathrm{C}}_1{\mathrm{C}}_2=\left|{\mathrm{r}}_1-{\mathrm{r}}_2\right| \end{gathered}$$

 one common tangent