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

Question

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?

Answer 1

Here is the picture for the problem, could not insert it as a comment, so it became an answer:

$5$ through the points $(2,-1)$, $(5,0)$, $(6,1)$, $(7,4)$, $(6,7)$, $(5,8)$, $(2,9)$, $(-1,8)$, $(-2,7)$, $(-3,4)$, $(-2,1)$, and $(-1,0)$">

In the picture, we see the circle with center $C(2,4)$ and radius $5$, it passes through the many points $(2,-1)$, $(5,0)$, $(6,1)$, $(7,4)$, $(6,7)$, $(5,8)$, $(2,9)$, $(-1,8)$, $(-2,7)$, $(-3,4)$, $(-2,1)$, and $(-1,0)$, and in the points $A(5,0)$ and $B(-1,8)$ we have a tangent with the same slope as the given line $3x-4y+k=0$. (Because such a line has the slope of $3x-4y=0$, or to have the diametral line passing through the center, $3x-4y=-10$, which also passes through the points $D(-2,1)$ and $E(6,7)$. Note that $DE\perp AB$. The parallel lines to $3x-4y=0$ going through $A$, respectively $B$, are $$ \begin{aligned} 3x-4y-15 &=0 &&\text{ through $A(5,0)$,}\\ 3x-4y+35 &=0 &&\text{ through $B(-1,8)$.} \end{aligned} $$

This is all that can be said mathematically.

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My speculation is that the "given line $3x-4y+\lambda$" (which is not an equation, well...), should have been the line with equation $3x-4y\color{red}{=}\lambda$. (There is also a comment with a red minus, it is the same story.) Then the numbers game applies, and we would have $a,b;\lambda$ to have the values either $5,0;15$, with the sum $20$, or...

Answer 2

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