Geometry – Position of a Point with Fixed Distance Between Two Points

Question

I have two points, $p_1$ and $p_2$, in a cartesian plane, and a fixed radius, $r$.

I want to find the coordinates of another point, $p_3$, that is in the same line of the $p_1$ and $p_2$, and always in a fixed distance, $r$, from the point $p_1$.

$(a,b)$       $(?,?)$                      $\quad \quad \quad\space(c,d)$

$p_1$———$p_3$———————$p_2$

        $\quad r$

Considering that the points $p_1$ and $p_2$ can be on anywhere in the plane.

Thanks in advance.

Answer 1

This is the problem of finding the point $p_3$$ (x,y)$ that divides the the joint of two points $p_1 (a,b)$ and $p_2 (c,d)$ internally in the ratio $r:t\normalsize$ (i.e. $p_1p_3:p_3p_1 = r:t$)

First you have to find the value of the consequent of the ratio i.e $t$, for that find the distance of $p_1p_2$ using this distance formula and then subtract $r$ from it.

Thus, $$ t = \left(\sqrt{ (c - a)^2 + (d-b)^2 } \right) -r $$

Now you can find $$ x= \frac{rc + ta}{r+t}$$ and $$ y=\frac{rd + tb}{r+t}$$