Line$ 3x-4y+k$ touches a circle $x^2+y^2-4x-8y-5 = 0$ at $(a,b)$, then find $k+a+b =?$

We can take advantage of the fact that the line touches the circle at a single point: $$y=\frac34x+\frac{\lambda}{4} \Rightarrow \\ x^2+\left(\frac34x+\frac{\lambda}{4}\right)^2-4x-8\left(\frac34x+\frac{\lambda}{4}\right)-5 = 0 \Rightarrow \\ 25x^2+2(3\lambda-80)x+\lambda^2-32\lambda-80=0 \quad (1)$$ The quadratic equation has a single solution when $D=0$: $$(3\lambda-80)^2-25(\lambda^2-32\lambda-80)=0 \Rightarrow \\ \lambda_1=-15,\lambda_2=35.$$ When we plug $\lambda=35>0$ into $(1)$: $$25x^2+50x+1225-1120-80=0 \Rightarrow \\ 25x^2+50x+25=0 \Rightarrow \\ x^2+2x+1=0 \Rightarrow \\ (x+1)^2=0 \Rightarrow \\ x=-1=a \Rightarrow y=8=b$$ Hence: $$\lambda+a+b=35-1+8=42.$$