I'm attempting to find the center of the purple circle (and/or the radius) given the following information:
- A point of tangency and the slope of the line (orange line)
- A point on a line that is tangent to the circle and the slope of the line (green line)
I cannot solve this trigonometrically because it is possible for the tangent lines to be at any angle to each other (i.e. including parallel). Edit: just to clarify, the diagram below is not intended to show the exact values, just to illustrate the information that is available (i.e. one tangent line with a point definitely on the circle, and another tangent line with a known point not on the circle). The tangent lines could be intersecting or parallel.
My intuition says I can use the equation of a circle and plug in the fixed tangency point and the equation of the other line to get a set of two equivalent circle equations, but I seem to be stuck there, and this is a bit out of the limit of my skillset (apologies if I'm making basic errors here):
$$(6 – h)^2 + (4 – k)^2 = r^2 = (1x – h)^2 + (0y – k)^2$$
Which seems to me to be equivalent to:
$$(6 – h)^2 + (4 – k)^2 = (1x – h)^2 + (0y – k)^2$$
But I'm not totally sure where to go from there.
Best Answer
The centers ($R$ and $S$ below) of the target circles lie on the parabola whose focus is the tangent point ($T$) and directrix is the "other" tangent line (through some point $U$).
As the figure shows, one readily determines that the corresponding radii ($r$ and $s$) satisfy $$r(1+\cos\theta) = d = s(1-\cos\theta) \quad\to\quad \{r,s\}=\frac{d}{1\pm\cos\theta}\tag{1}$$ where $d$ is the focus-directrix distance (ie, the distance from the tangent point to the "other" tangent line) and $\theta$ is the angle between direction vectors of the two lines (equivalently, between normal vectors).
Now, if $t$ and $u$ are the unit normal vectors to the tangent lines through $T$ and $U$, then we can write $$d = |(U-T)\cdot u| \qquad \cos\theta = t\cdot u \tag2$$ so that $$\{R,S\} = T \pm_1 \frac{|(U-T)\cdot u|}{1\pm t\cdot u} \;t \tag3$$ where $\pm_1$ is a hedge on my part in case of a sign error due to reversing a vector somewhere. Barring an error, you "should" be able to leverage the signed distance $d$ to write $$\{R,S\} = T \pm \frac{(U-T)\cdot u}{1\pm t\cdot u}\;t \tag4$$ where the two $\pm$s match.
(Sanity check: Changing the direction of $t$ effectively reverses each $\pm$, so that's consistent. Likewise for $u$. So the only question is whether I should've written $T-U$ instead of $U-T$, but I believe my GeoGebra sketch bears me out on that one.)
Note that, if the lines are parallel, then $t\cdot u = \pm 1$ (with the sign depending upon how the normals are chosen), so that one of the centers $R$, $S$ lies halfway between the lines, and the other shoots off to infinity, as expected. $\square$