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I have a circle with radius r and center $(c_x, c_y)$. I have a line segment $(x_1, y_1)$ and $(x_2, y_2)$ given $(x_2, y_2)$ is always a point inside the circle.

I am trying to find the intersection between the circle and the line segment. I have tried the tangent formula that mentioned here i am not sure if it works in my case.

enter image description here

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  • $\begingroup$ I'm assuming (x1,y1) is outside of the circle? $\endgroup$ – Bolun Zhang Aug 12 '14 at 0:38
  • $\begingroup$ @3.141592653589793 Yes... $\endgroup$ – nida8191 Aug 12 '14 at 15:10
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The equation of the circle is:

$$(x-c_x)^2+(y-c_y)^2=r^2$$

The equation of the line is:

$$y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1) \ \ \ (*)$$

To find the intersection solve at the $(*)$ for $y$ and replace it at the equation of the circle.

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Use parametrization of segment. This function describes each point on segment for $t \in [0,1]$

$$f(t)=(x_1,y_1)+t((x_2,y_2)-(x_1,y_1))=(t_1,t_2)$$

Now you now how to calculate distance $(x,y)$ from $(c_x,c_y)$, it's:

$$dist(x,y)=\sqrt{(c_x-x)^2+(c_y-y)^2}$$

So you are looking for such a $t$, for which

$$dist(f(t))=r$$

Where $r$-radius. If you substitute $f(t)=(x_1,y_1)+t((x_2,y_2)-(x_1,y_1))=(t_1,t_2)$ you get quadratic equation, next you calculate $t$ and use $t$ to get intersection (you can get 0,1 or two solutions).

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WLOG, the center of the circle is the origin (otherwise, translate all points to make it such).

And WLOG, the length of the segment is $1$ (otherwise, divide all coordinates by that length).

Plug the parametric equation of the line segment in the implicit equation of the circle,

$$x^2+y^2=(x_1+t\delta x)^2+(y_1+t\delta y)^2=r^2.$$

The solving the quadratic equation,

$$t=-(x_1\delta x+y_1\delta y)\pm\sqrt{(x_1\delta x+y_1\delta y)^2-(x_1^2+y_1^2-r^2)}.$$

The rest is easy. (don't forget to undo the transformations.)

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WLOG, the center of the circle is the origin (otherwise, translate all points to make it such).

Now scale the circle so that the radius is $1$ (not the line segment as Yves Daoust did!).

Let the line by given with $y=mx+c$ where $m=(y_2-y_1)/(x_2-x_1)$ and $c=y_1 - mx_1$ then if $$\begin{align}m^2 + 1 - c^2 > 0\end{align}$$ the line intersects the circle, if $$\begin{align}m^2 + 1 - c^2 < 0\end{align}$$ it does not intersect and if $$\begin{align}m^2 + 1 - c^2 = 0\end{align}$$ it touches it in one point.

This is a special case of the problem of intersecting a line with an ellipse.

As is given on that page the coordinates of intersection (if they exist) are $$\begin{align}x_{1,2}=&\frac{-mc\pm\sqrt{m^2+1-c^2}}{m^2+1},&y_{1,2}=&\frac{c\pm m\sqrt{m^2+1-c^2}}{m^2+1}\\\end{align}$$

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