Line$ 3x-4y+k$ touches a circle $x^2+y^2-4x-8y-5 = 0$ at $(a,b)$, then find $k+a+b =?$ 
The line $3x-4y+\lambda=0, (\lambda > 0)$ touches the circle $x^2+y^2-4x-8y-5 = 0$ at $(a,b)$. Find the value of $\lambda+a+b$.

I tried solving in the following manner:  
Clearly the circle has center at $(2,4)$ with a radius of $5$.
So, the perpendicular distance from the center onto the given line must be the radius (Radius is perpendicular to tangent)
So, using the formula for perpendicular distance
$$\frac{|3\times2-4\times4+\lambda|}{(3^2+4^2)^{1/2}}= 5$$
$$\implies\frac{|3\times2-4\times4+\lambda|}{5} = 5$$
$$\implies|\lambda-10| = 25$$
From this, I get $\lambda = 35,-15$, but since in the question it is given $\lambda>0$, I'm taking $\lambda  = 35$
Now, the line is $3x-4y+35 = 0$
Given, it is the tangent at $(a,b)$
Writing equation of tangent at (a,b), I get, 
$$ax+by+(-2)(x+a)+(-4)(y+b)-5 = 0$$
$$\implies (a-2)x+(b-4)y+-2a-4b-5 = 0$$
which should be identical to the line $$3x-4y+35= 0$$
So, I write
$$\frac{(a-2)}{3} = \frac{b-4}{-4}=\frac{-(2a+4b+5)}{35}$$
Solving these, I obtained that a = -1 and b = 8  
So $\lambda+a+b = 42$
But the answer says 20. Is there a problem with my above procedure?

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:
$$\lambda+a+b=35-1+8=42.$$ 
