I have this situation (forgive my paint skills):

I have the radius R, the x coordinate D, and the angle of the tangent a. How would I go about finding h, the y coordinate of that point?
Thanks!
I have this situation (forgive my paint skills):

I have the radius R, the x coordinate D, and the angle of the tangent a. How would I go about finding h, the y coordinate of that point?
Thanks!
By the Pythaogrean theorem
$$x^2+y^2=r^2$$
Note that $$2x+2yy'=0\implies y'=-\frac{x}{y}$$
which means that if your slope provided is positive, $y$ and $x$ are of different signs.
Examining the figure below, from triangle $OPQ$, we have
$R=|OP|+|h|$
Using trigonometry, $|OP|=R\sin(90^{\circ}-a)=R\cos(a)$
Therefore
$|h|=R-|OP|=R-R\cos(a)=R(1-\cos(a))$
Assuming that the center of the circle is the origin $(0,0)$, the coordinate of point $Q$ will be $(D,-R\cos(a))$
