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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 $p_1$.

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

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

Thanks in advance.

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up vote 1 down vote accepted

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 $$ y=\frac{rd + tb}{r+t}$$

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The denominator should be just $t$ – Tapu Nov 25 '11 at 19:08
Sorry, I have removed the down vote now...plz edit your post :) – Tapu Nov 25 '11 at 19:12
@Swapan: Fixed. – Quixotic Nov 25 '11 at 19:14

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.

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This is on the line between $p_1$ and $p_2$ as long as $|p_2-p_1|\ge r$. – robjohn Nov 25 '11 at 18:34

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 $|P_1P_2|=\sqrt{(a-c)^2+(b-d)^2}$

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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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are you sure about the denominator? – Quixotic Nov 25 '11 at 18:56
@MaX You may wish to edit your answer once more to correct present $t$ to $t-r$, or to put the denominator as just $t$ :) – Tapu Nov 25 '11 at 19:06
@MaX No problem....but please be patient, I never argue (and sometime bet) without being sure that "If you can prove me false, it will imply 0=1" :D – Tapu Nov 25 '11 at 19:15

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