Say I have the coordinates of a point named $A$ on a linear line, the distance from point $A$, and the equation of the line, how do I find point $B$ and $C$?
I figured that the equation of a circle could possibly help in finding both points with the intersection of the circle and the line - with the distance being the radius, and point $A$ being the center of the circle, and then have a system of equations with both the circle equation and the linear line equation, but then I end up with two variables, which is beyond my ability to solve.
If the slope is zero or infinity, then the line is either the $x$ or the $y$ axis, so in that case this question is easy. We will deal with a non-zero non-infinite slope.
If $d$ be the given distance, then according to the given diagram:
If the slope of the line is $L$, then we know that $\tan^{-1} L = \theta$, the angle that the line makes with the $x$ axis. I have drawn two lines which are parallel to the $x$-axis below. It can be seen from the diagram, that the $x$ coordinate of $C$ is $d \sin \theta$ below that of $A$, and the $y$ coordinate is $d \sin \theta$ below that of $A$. So if $A = (a,b)$, then $C = (a-d\cos \theta, b - d \sin \theta)$.
Now, if $L = \tan \theta$, then we can use ordinary trigonometric formulas to conclude that $\sin \theta = \frac{L}{\sqrt{L^2+1}}$, and $\cos \theta = \frac{1}{\sqrt{L^2+1}}$.
Therefore, if $A = (a,b)$ is the point and $d$ is the given distance, then: $$ \bbox[yellow,5px, border:2px solid red ] {C = \left(a - \frac{d}{\sqrt{L^2+1}}, b-\frac{dL}{\sqrt{L^2+1}}\right)} $$
The formula for $B$ is just the same, but with plus signs.
$$ \bbox[yellow,5px, border:2px solid red ] {B = \left(a + \frac{d}{\sqrt{L^2+1}}, b+\frac{dL}{\sqrt{L^2+1}}\right)} $$