<s>[INST] 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 $k$?
 [/INST] If the line is a tangent to the circle, it means that the line and circle have one point in common.
Substituting $y = kx + 1$ into the equation of the circle gives $9 = (kx)^2 + (x-5)^2$, which simplifies to $0 = (k^2+1)x^2 - 10x + 16$.
The discriminant of the quadratic must be zero:
$\Delta = (-10)^2 -4(k^2+1)(16) \rightarrow k = \pm \frac{3}{4}$.
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