The point where the line 4x – 3y + 7 = 0 touches the circle $x^2+y^2-6x+4y-12=0$

It is given that 4x−3y+7=0 is a tangent to the circle x2+y2−6x+4y−12=0. 

The centre of the circle C is (3,−2). 

The line will touch the circle at the foot of the perpendicular from the centre

Foot of the perpendicular from C to tangent is 
$$\begin{gathered} \frac{h-3}{4}=\frac{k+2}{-3}=\frac{-\left(4\right(3)-3(-2)+7)}{{\left(4\right)}^2+{(-3)}^2} \\ \Rightarrow \frac{h-3}{4}=\frac{k+2}{-3}=-1 \\ \Rightarrow \frac{h-3}{4}==-1 and \frac{k+2}{-3}=-1 \\ \Rightarrow h=-1,k=1 \end{gathered}$$

Hence the point of contact is (−1,1)