From 6255d768ba8e76066c5f44603674c1199dcda4af Mon Sep 17 00:00:00 2001 From: Luca Leone Date: Wed, 1 Sep 2021 00:16:11 +0200 Subject: [PATCH] Update content.en-GB.md (#327) Fix typo --- docs/chapters/pointcurves/content.en-GB.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/docs/chapters/pointcurves/content.en-GB.md b/docs/chapters/pointcurves/content.en-GB.md index b4035ca3..6d2f9d8c 100644 --- a/docs/chapters/pointcurves/content.en-GB.md +++ b/docs/chapters/pointcurves/content.en-GB.md @@ -18,7 +18,7 @@ With this code in place, creating a quadratic curve from three points is literal For cubic curves we need to do a little more work, but really only just a little. We're first going to assume that a decent curve through the three points should approximate a circular arc, which first requires knowing how to fit a circle to three points. You may remember (if you ever learned it!) that a line between two points on a circle is called a [chord](https://en.wikipedia.org/wiki/Chord_%28geometry%29), and that one property of chords is that the line from the center of any chord, perpendicular to that chord, passes through the center of the circle. -That means that if we have have three points on a circle, we have three (different) chords, and consequently, three (different) lines that go from those chords through the center of the circle: if we find two of those lines, then their intersection will be our circle's center, and the circle's radius will—by definition!—be the distance from the center to any of our three points: +That means that if we have three points on a circle, we have three (different) chords, and consequently, three (different) lines that go from those chords through the center of the circle: if we find two of those lines, then their intersection will be our circle's center, and the circle's radius will—by definition!—be the distance from the center to any of our three points: