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First, imagine we have a circle. And imagine we have a line that extends from the center of the circle in one direction. At the end of that line is point B. The center of the circle is point A. How do I get a point on the circumference which is touching the line? Here is an image to explain it better: enter image description here

Sorry if my image wasn't good, Thanks for any help.

Rishi
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  • What do you mean by "get a point". If you mean construct it, in the sense of compass and straightedge, then it is already constructed (it's the intersection of the ray and the circle). If you want coordinates for it, you'll need coordinate expressions for everything else. Or did you mean something else? – lulu Aug 06 '17 at 15:31
  • I mean coordinates like x and y – Rishi Aug 06 '17 at 15:38
  • Ok, but then you'll need coordinates for everything else. Say the center is $(0,0)$ and the radius is $r$ and $B=(a,a)$. Then the ray is $y=x$ for $x≥0$, the circle is $x^2+y^2=r^2$ and the intersection, $P$ is given by $x^2+x^2=r^2,,r≥0\implies P=(\frac r{\sqrt 2},\frac r{\sqrt 2})$. Of course you might have different coordinates, but the method will be the same. – lulu Aug 06 '17 at 15:45
  • Thanks for the help, post it as a answer so I can accept it – Rishi Aug 06 '17 at 15:48

3 Answers3

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To get an answer in coordinates, you'll need coordinates for everything else in the picture.

As an example to illustrate the method: Say the center is $(0,0)$ and the radius is $r$ and $B=(a,a)$. Then the ray is $y=x$ for $x≥0$, the circle is $x^2+y^2=r^2$ and the intersection, $P$ is given by $$x^2+x^2=r^2,\,r≥0\implies P=(\frac r{\sqrt 2},\frac r{\sqrt 2})$$ Of course you might have different coordinates, but the method will be the same.

lulu
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Let $A = (x_1,y_1)$ and $B = (x_2,y_2)$, and suppose that the radius of the circle is $r$ (where, I suppose, we should assume that $r$ is less than the distance from $A$ to $B$). The equation of a line through a point $(x_1,y_1)$ with a given slope $m$ is $$ y - y_1 = m(x - x_1),$$ and we can compute the slope of the line from $A$ to $B$ as $$ m = \frac{y_1 - y_2}{x_1 - x_2}. $$ Thus the point of intersection lies on the line $$ y - y_1 = \frac{y_1-y_2}{x_1-x_2}(x-x_1). $$ Note that if you actually know what the coordinates of $A$ and $B$ are, this can be significantly simplified---everything in that equation is a constant other than $x$ and $y$.

Next, you know that the point of intersection is on the circle centered at $A$. The equation of the circle is given by $$ (x-x_1)^2 + (y-y_1)^2 = r^2. $$ Again, if you actually know $x_1$, $y_1$, and $r$ this can be simplified quite a bit.

Since the point of intersection must be on both the line and the circle, we must solve the system $$ \begin{cases} y - y_1 = \dfrac{y_1-y_2}{x_1-x_2}(x-x_1) \\ (x-x_1)^2 + (y-y_1)^2 = r^2. \end{cases} $$ This system can be solved by solving the first equation for $y$ and substituting that into the second equation. The second equation can then be solved for $x$ (it is quadratic in $x$, so that should be relatively straight-forward, assuming that you have at least taken some algebra). Finally, substitute that back into the first equation to get $y$. The point of intersection will be one of the two solutions to the system.

Xander Henderson
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Assume that point $A$ is at the origin and that the circle has radius $r$. Furthermore, let the point $B=(X,Y)$. Then the desired point, say $P(x,y)$ is given by

$$ P=(r\cos \theta,r\sin \theta)\\ \theta=\tan^{-1}\frac{Y}{X} $$

EDIT: Another solution comes to mind. Let $R=\sqrt{X^2+Y^2}$, then

$$P(x,y)=\frac{r}{R}B(X,Y)$$

Cye Waldman
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