The equation of one of the tangents from (1, 1) to a circle with its centre at (3, 0) is 3x + y – 4 = 0. The equation of the other tangent is ...........

radius= $\perp \mathrm{distance} \mathrm{from} \mathrm{centre}\left(3,0\right) \mathrm{to} \mathrm{the} \mathrm{tangent} 3\mathrm{x}+\mathrm{y}-4=0$

$\mathrm{r}=\frac{3\left(3\right)+0-4}{\sqrt{9+1}}=\frac{5}{\sqrt{10}}$.  other tangent be y –1 = m(x–1)$\Rightarrow mx-y+(1-m)=0$

 $$\begin{gathered} \mathrm{r}=\mathrm{d}\Rightarrow \frac{5}{\sqrt{10}}=\frac{3\mathrm{m}+0+(1-\mathrm{m})}{\sqrt{{\mathrm{m}}^2+1}} \\ \Rightarrow 25(1+{\mathrm{m}}^2)=10{\left(2\mathrm{m}+1\right)}^2 \\ \Rightarrow 15{\mathrm{m}}^2+40\mathrm{m}-15=0 \\ \therefore 3{\mathrm{m}}^2+8\mathrm{m}-3=0 \\ \Rightarrow \mathrm{m}=\frac{1}{3}\mathrm{substituting} \mathrm{m}=1/3 \mathrm{in} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{tangent} \\ \mathrm{y}-1=\frac{1}{3}\left(\mathrm{x}-1\right)\Rightarrow \mathrm{x}-3\mathrm{y}+2=0 \end{gathered}$$