I have an app, where I draw a graph. From each circle in this graph there are some lines to other circles. To not mess up my drawing, I tend to draw the line up from the circle, then horizontally in direction of the other circle. Then I stop some pixels before the other circle center and I draw the line from there to the other circles center. Now I wanted to clear the drawing, cuz when there are many lines going to the same circle, it's hard to see if the line ends with an arrow.

So I thought the lines would go into circles center direction, but they would stop at the circumference. Here's how it looks like:

enter image description here

As you can see, the lines which come from the left side to the right side (f.e from q1 to q2) are really fine, definitely pointing at the center, but not going inside.

But what about the lines coming from the right to the left (f.e from q4 to q1 or q3 to q1). You can clearly see, that they stop at the circumference, but they definitely do not point at the center, which is really not aesthetic.

This is the algorithm I came up with:

1.  I have X and Y of the point from I will be drawing the line (the final line, cuz every
    line consists of 3 lines -> the one which goes up or down, the one which goes
    horizontally and the last one, which connects the end of the 2nd line with circle center)
2.  Then I take the X and Y of the circle to which I would be drawing a line
3.  a = circleCenter.X - lineEnd.X
    b = circleCenter.Y - lineEnd.Y
    c = sqrt(a^2 + b^2)
4.  sin(alfa) = b/c
5.  alfa = asin(sin(alfa)) - PI
6.  Now I want to get the point on the circle:
    newPoint.X = R * cos(alfa)
    newPoint.Y = R * sin(alfa)
7.  newPoint has X and Y like the circle would be in 0, 0, so I need to do:
    newPoint.X = circleCenter.X + newPoint.X
    newPoint.Y = circleCenter.Y + newPoint.Y
8. And then I draw to this point

And as you can see, it works perfectly for those coming from the left, but not so well for those coming from the right

  • $\begingroup$ for lines coming from the right you do not need to subtract $\pi$ $\endgroup$
    – Vasili
    Commented Oct 24, 2018 at 19:46
  • $\begingroup$ If I do what you say - subtract PI if coming from the left, and not subtract when coming from the right, this happens: i.imgur.com/DjBAF64.png $\endgroup$ Commented Oct 24, 2018 at 19:58
  • $\begingroup$ your problem is with angle calculations: you have to consider 4 cases: right line coming from top or bottom, left line coming from top or bottom. Depending on that, cosine and sine function will have a different sign. When you take asin, it gives you angle between $-\pi/2$ and $\pi/2$ $\endgroup$
    – Vasili
    Commented Oct 24, 2018 at 20:11

1 Answer 1


The asin function requires you to do various tricks to deal with lines that can come from any direction (upper right, upper left, lower right, lower left) because it only gives you half a circle's worth of angles, $-\frac\pi2$ to $\frac\pi2,$ whereas you need angles all around the circle.

The atan2 function, if your software library has it, is usually much better for applications like this. You call it like this:

    atan2(b, a)

and it gives you a full range of angles from $-\pi$ to $\pi,$ which will be sufficient for lines coming from any direction. Moreover, you don't even need to compute $c.$

  • $\begingroup$ Wow, that really helps, now all the lines are actually pointing at the center. The problem is, the points given by those angles tend to make lines go through the circle: i.imgur.com/c07B1WT.png. Is there a way around it? $\endgroup$ Commented Oct 24, 2018 at 20:37
  • $\begingroup$ It looks like every single line went all the way across the circle. If you're still subtracting PI, then you should be able to just stop doing that and it will work. Otherwise you can see the point is in the exact opposite direction from the center you want, so do the opposite of whatever you were doing, e.g., subtract instead of adding. $\endgroup$
    – David K
    Commented Oct 24, 2018 at 20:46
  • 1
    $\begingroup$ I think you need to use atan2(-b,-a) $\endgroup$
    – Vasili
    Commented Oct 24, 2018 at 20:48
  • $\begingroup$ @DavidK I am not subtracting PI anymore. @Vasya, this is it! atan2(-b, -a) works like a charm! $\endgroup$ Commented Oct 24, 2018 at 21:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .