Position of a point with fixed distance between other two points
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.
$\endgroup$4 Answers
$\begingroup$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}$$
$\endgroup$ 3 $\begingroup$Do you mean $$ 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|} $$ works also.
$\endgroup$ 1 $\begingroup$Let's define point $P_3$ as $P_3(e,f)$ .
You may calculate coordinates $e$ and $f$ using following equalities :
$e=\frac{a+\lambda c}{1+\lambda}$ , and $f= \frac{b+\lambda d}{1+\lambda}$
where $\lambda=\frac{r}{|P_1P_2|-r}$ , and $|P_1P_2|=\sqrt{(a-c)^2+(b-d)^2}$
$\endgroup$ $\begingroup$Are you sure $P_3$ lies inside the segment $P_1P_2$? This means $r\le \sqrt{(a-c)^2+(b-d)^2}$. In this case the point $P_3$ is given by $$\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)$$
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