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 $C$?

enter image description here

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:enter image description here

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.

$$ \bbox[yellow,5px, border:2px solid red ] {B = \left(a + \frac{d}{\sqrt{L^2+1}}, b+\frac{dL}{\sqrt{L^2+1}}\right)} $$

  • $\begingroup$ Thank you as well for your very detailed answer. $\endgroup$ – RandomUser Feb 2 '18 at 0:02
  • $\begingroup$ You are welcome! $\endgroup$ – астон вілла олоф мэллбэрг Feb 2 '18 at 0:05
  • $\begingroup$ Is there a single formula that encompasses all 3 of them? $\endgroup$ – RandomUser Feb 2 '18 at 15:24
  • $\begingroup$ Do you want a single formula for both $B$ and $C$? Which three are you referring to, because I highlighted only two of them. $\endgroup$ – астон вілла олоф мэллбэрг Feb 3 '18 at 2:49
  • $\begingroup$ I mean all 3 types of linear lines, with zero slope, infinite slope, and a slope that is non-zero and non-infinite $\endgroup$ – RandomUser Feb 4 '18 at 3:51

Let $AB=AC=k$, and the slope of the line be $m$.

Then we know that AB's x-coordinate difference is $\frac{k}{\sqrt{m^2+1^2}}$ and y-coordinate difference is $\frac{k}{\sqrt{m^2+1^2}}$.

Do you mean to find this? (Formula above are from Pyth. thm.)

Remark: if you use equations to solve, after a substitution of the line equation into circle equation, you will end up with quadratic equation.

  • $\begingroup$ Yes, thank you very much, this is what I meant. $\endgroup$ – RandomUser Feb 2 '18 at 0:05

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