Position of a point with fixed distance between other 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 $$ y=\frac{rd + tb}{r+t}$$

Answer 2

Do you mean $$ p_1+r\frac{p_2-p_1}{|p_2-p_1|} $$ Afterthought:

If you're not worried about being on the line segment between $p_1$ and $p_2$, then $$ p_1-r\frac{p_2-p_1}{|p_2-p_1|} $$ works also.

Answer 3

Are you sure $P_3$ lies inside the segment $P_1P_2$? This means $r\le \sqrt{(a-c)^2+(b-d)^2}$. In this case the point $P_3$ is given by $$\left(\frac{a \times (\sqrt{(a-c)^2+(b-d)^2}-r)+c \times r }{\sqrt{(a-c)^2+(b-d)^2}},\frac{b \times (\sqrt{(a-c)^2+(b-d)^2}-r)+d \times r}{\sqrt{(a-c)^2+(b-d)^2}}\right)$$

Answer 4

Let's define point $P_3$ as $P_3(e,f)$ .

You may calculate coordinates $e$ and $f$ using following equalities :

$e=\frac{a+\lambda c}{1+\lambda}$ , and $f= \frac{b+\lambda d}{1+\lambda}$

where $\lambda=\frac{r}{|P_1P_2|-r}$ , and $|P_1P_2|=\sqrt{(a-c)^2+(b-d)^2}$