Given the circumference $(x-3)^2+(y-2)^2=13$ find $k$ where $k$ is a coefficient in the parametric equation $(k+1)x+8ky-6k+2=0$ of the lines passing through the points $A(0;4)$, $B(6;4)$, $C(1;-1)$. Here's how I did it.
I found the equations for the lines:
$a: -3x+2y=8$ (passing through A)
$b: -3x+8y=14$ (passing through B)
$c: 2x+3y=-1$ (passing through C)
Now here's the problem, how do I express $a$, $b$ and $c$ in $(k+1)x+8ky-6k+2=0$ form and find $k$?
What you want to do is multiply both sides of (a), (b) and (c) equations by $n$ and compare them to $(k+1)x+8ky-6k+2$.
For example in equation (a) multiply both sides of $-3x+2y=8$ by $n$. Notice that the linear equation is still the exact same.
$$(-3n)x+(2n)y=8n$$
Now compare this to $(k+1)x+(8k)y=6k-2$
We get
(1) $-3n=k+1$
(2) $2n=8k$
(3)$8n=6k-2$
You can just solve equations (1) and (2). I got $n=-4/13$ and $k=-1/13$. To check the solutions; substitute $n$ and $k$ in (1), (2) and (3). Apparently the solution works for all of them.
Now repeat this with (b) and (c), the other two equation in your post and solve for their values of $k$.