Equation of the tangent to the circle $x^2+y^2-2x+4y-4=0$ which is parallel to the line $3x+4y-1=0$ is 

Let equation of tangent parallel to given line is 

$$\begin{gathered} 3\mathrm{x}+4\mathrm{y}+\mathrm{k}=0\to \left(1\right) \\ Given circle x^2+y^2-2x+4y-4=0 \\ centre \left(C\right)=(1, -2) and radius \\ \mathrm{r}=\sqrt{1+4+4}=3 \\ Since \left(1\right) is \tan gent then \\ \mathrm{r}=\mathrm{d} \\ \Rightarrow 3=\frac{\left|3-8+\mathrm{k}\right|}{\sqrt{8+16}} \\ \Rightarrow \left|-5+\mathrm{k}\right|=15 \\ \Rightarrow \mathrm{k}-5=\pm 15 \\ \mathrm{k}=-10, \mathrm{k}=20 \\ \Rightarrow 3\mathrm{x}+4\mathrm{y}-10=0 \end{gathered}$$