# Determine the closest point along a circle's $(x_1, y_1)$ radius from any point $(x_2, y_2)$, inside or outside the radius of the circle.

I have a circle centered at point $(x_1, y_1)$ and another point at $(x_2, y_2)$. This point, $(x_2, y_2)$ may or may not be within the radius ($r$) of the circle. I wanted to create a line going from the center of the circle $(x_1, y_1)$ to the point $(x_2, y_2)$. I have been using the quadratic formula to determine the intersection between this line and the radius of the circle. In my application, the point $(x_2, y_2)$ was always outside of the circle, so it made sense to always have a single intersection point along the radius of the circle. However, when $(x_2, y_2)$ is within the circle, I was not able to determine a point of intersection.

Is there a simple way to determine the closest point from $(x_2, y_2)$ to the circle's radius? If the line from $(x_1, y_1)$ were to extend forever, it would be the intersecting point of this extended line the circle's radius.

I am using the quadratic equation, but I am having troubles figuring out the intersection, without artificially extending the line. Is it also possible to determine the closest point from $(x_2, y_2)$ to the circle's radius if you don't know whether it is in or out of the radius? If wondering, I am using this in a coding environment.

I added my Inkscape drawing (editing with Paint it to show my application to match the question; please ignore the missing line and resulting gap from the circle).

Thank you.

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You must learn LaTeX markup! –  kjetil b halvorsen Mar 28 at 19:34
Thanks. I was not sure that it could be used here. –  baconcow Mar 29 at 17:05

I eventually found this answer here. Might be best to mark as duplicate, as this solution worked for me. Here is a summary of the solution as required within my application.

Point of intersection along the circumference of a circle centered at $(x_1, y_1)$ with a radius, $r$.

$x_{int} = x_1+\frac{r (x_2-x_1)}{\sqrt{(x_2-x_1)+(y_2-y_1)}}$

$y_{int} = y_1+\frac{r (y_2-y_1)}{\sqrt{(x_2-x_1)+(y_2-y_1)}}$

Where $(x_{int}, y_{int})$ is the point of intersection along the circumference of the circle with range $r$.

This apparently works for both scenarios where point $(x_2, y_2)$ is either within or outside of the radius of the circle.

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