1. Question about tangent lines to circles...

I'm using the Sullivan Precalculus 7th Edition book for self-study. I actually have a background all the way through integral calculus, but that was years ago and I'm trying to relearn much of what I've forgotten. In particular, one of the early problems has me completely stumped and I've tried every way I can think of to morph the equations to get the result it wants. The solution in the back of the book is useless because I do not know where the author came up with the initial equations.

The question is as follows: The tangent line to a circle may be defined as the line that intersects the circle in a single point, called the point of tangency (this I knew). if the equation of the circle is $x^2 + y^2 = r^2$ and the equation of the tangent line is y = mx + b, show that:
(a) $r^2(1 + m^2) = b^2$
Hint: The quadratic equation $x^2 + (mx + b)^2 = r^2$ has exactly one solution.

I was able to use the discriminant to solve this part of the problem.

(b) The point of tangency is $( - (r^2m)/b, r^2/b)$.

This is where I'm stuck. I tried substituting $r^2(1 + m^2)$ for $b^2$ into the equation for the intersection between the circle and the line $x^2(1 + m^2) + 2bmx + b^2 = r^2$ in an attempt to find x, but no matter how I manipulate the variables, I cannot seem to get it to work. Every time I wind up with some completely unwieldy equation that I cannot do anything more with. I hate having problems I can't solve, and I'm hoping one of you here might be able to help me figure out how to get the result.

2. Never mind... after spending many hours on the problem I finally figured out that I can use the quadratic formula to solve for x, and then use that result to substitute into y = mx + b to find y.

3. Use the point of tangency and the distance formula:

$\left(\frac{-r^2m}{b}\right)^2 + \left(\frac{r^2}{b}\right)^2 = r^2$

$\frac{r^4m^2}{b^2} + \frac{r^4}{b^2} = r^2$

Multiply both sides by $\frac{b^2}{r^2}$:

$r^2m^2 + r^2 = b^2$

$r^2(m^2 + 1) = b^2$