Well, I was testing somethings out to try and find a particular point that is a tangent to a circle and I decided to try this. I'm not too sure whether it's right or not.

I just want to know if this is true for all tangents to a circle?

Excuse the terrible picture. I just drew it in a rush on MS Paint. (Rofl) I am hoping you can decipher it.

You've found a very special case which is only true if \(\displaystyle |\overline{CR}| = |\overline{CQ}|\)

In general:

1. Use the indicated blue triangle to determine the slope of the tangent:

\(\displaystyle m_t=-\frac1{m_{radius}}~\implies~m_t=-\frac{X_P - X_C}{Y_P - Y_C}\)

2. Use the triangle RCQ to determine the slope of the tangent:

\(\displaystyle m_t=\frac{Y_Q-Y_R}{X_Q-X_R}\)

Since \(\displaystyle X_R = X_C\) and \(\displaystyle Y_Q = 0\) the 2nd equation becomes:

\(\displaystyle m_t=\frac{-Y_R}{X_Q-X_C}\)

3. You now can use these 2 equations to get some additional results.