Calculus A Level Line Tangent to Circle How can you find values of $k$ such  that $y = kx + 1$ is tangent to the circle $(y-1)^2 + (x-5)^2 = 9 $?
I first rewrote the circle equation in terms of y:
$$ (y-1)^2 = -(x-5)^2 + 9 \\y-1 = \pm\sqrt{-x^2+10x -16} \\ y = \pm\sqrt{-x^2+10x -16} + 1 $$
Then I took two derivatives, one for the positive equation and one for the negative equation:
$$y\prime_{_{+}} = \frac{1}{2}(-x^2+10x-16)^\frac{-1}{2}(-2x+10) \\=-(x-5) (-x^2+10x-16)^\frac{-1}{2}\\y\prime_{_{-}} = -\frac{1}{2}(-x^2+10x-16)^\frac{-1}{2}(-2x+10) \\=(x-5) (-x^2+10x-16)^\frac{-1}{2}$$ 
So if I look at just the positive semicircle, I have a line $y = kx + 1$ which intersects $y = \sqrt{-x^2+10x-16} + 1 $ at one spot and tangent at that spot $ k = -(x-5)(-x^2+10x-16)^\frac{-1}{2} $.
How do I actually find the value of $k$?

You can use implicit differentiation to find the slope $y'$ at any $(x,y) $ in the circle with $y \neq 1$: 
$$ 2(y-1)y' + 2(x-5) = 0 \implies y' = \dfrac{5-x}{y-1} $$
if $y' = k$, then we have $5-x = k(y-1)$ and together with the equation of the circle and the equation of the line, we obtain the following system of three equations and three unknowns: 
$$ \begin{cases} 5-x = k(y-1) \\ y = kx + 1 \\ (y-1)^2 + (x-5)^2 = 9 \end{cases} $$
Solving this will find your k
