Calculus A Level Line Tangent to Circle

Question

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 $k$?

Answer 1

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

Answer 2

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:

$\Delta = (-10)^2 -4(k^2+1)(16) \rightarrow k = \pm \frac{3}{4}$.