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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 linea tangent to the circle with center
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, which is the slope of our tangent line. Similarly, there will be a symmetric tangent line with a slope of
.
Their solution:
...(algebraic work that I can follow)...
Where is their starting equation coming from? Is this some distance from a line to a point equation that I've never seen?
Originally Posted by Henderson 
