To begin with, I solved this problem, and got the correct answer. The reason I'm posting is because the solution manual uses a different method, and I can't follow what they're doing. Please explain their solution.

Problem:

For what values of k is the straight line $\displaystyle y = kx + 1$ a tangent to the circle with center $\displaystyle (5,1)$ and radius 3?

My solution:

The distance from the y-intercept (0,1) to the center of the circle is 5, and the radius is 3. Since the tangent is perpendicular to the radius, a 3-4-5 triangle is created. Dropping an altitude in this triangle gives us a change-in-y, change-in-x triangle for the tangent line. Since it is similar to the large triangle, it also has an opposite/adjacent ratio of $\displaystyle \frac{3}{4}$, which is the slope of our tangent line. Similarly, there will be a symmetric tangent line with a slope of $\displaystyle \frac{-3}{4}$.

Their solution:

$\displaystyle

\frac{|5k-1+1|}{\sqrt{k^2+1}} = 3

$

...(algebraic work that I can follow)...

$\displaystyle

k = \pm\frac{3}{4}

$

Where is their starting equation coming from? Is this some distance from a line to a point equation that I've never seen?