Equation of a Circle, given three points
Problem: Give the general solution to this question:
Find a circle given 3 points (a,b), (c,d), and (e,f)
Your answer should be in the form of an equation of the circle centered at point (h,k) with radius r, and should look like this:
(x-h)² + (y-k)² = r²
Solve for h, k, and r in terms of a, b, c, d, e, and f.
Solution:
Start with these three equations:
(a-h)² + (b-k)² = r²
(c-h)² + (d-k)² = r²
(e-h)² + (f-k)² = r²
Multiply them out, and you get these three equivalent equations:
a² - 2ah + h² + b² - 2bk + k² = r²
c² - 2ch + h² + d² - 2dk + k² = r²
e² - 2eh + h² + f² - 2fk + k² = r²
Multiply the first by (e-c), the second by (a-e), and the third by (c-a):
(a² - 2ah + h² + b² - 2bk + k²)(e-c) = r²(e-c)
(c² - 2ch + h² + d² - 2dk + k²)(a-e) = r²(a-e)
(e² - 2eh + h² + f² - 2fk + k²)(c-a) = r²(c-a)
Multiply out these three equations, then add 'em up. Amazingly, all the
squared h, k, and r terms go away, so you can easily solve for h and k
a²e - a²c - 2aeh + 2ach + h²e - h²c + b²e - b²c - 2bek + 2bck + k²e - k²c = r²e - r²c
c²a - c²e - 2cah + 2ceh + h²a - h²e + d²a - d²e - 2dak + 2dek + k²a - k²e = r²a - r²e
e²c - e²a - 2ech + 2eah + h²c - h²a + f²c - f²a - 2fck + 2fak + k²c - k²a = r²c - r²a
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a²e - a²c + c²a - c²e + e²c - e²a + b²e - b²c + d²a - d²e + f²c - f²a -2bek + 2bck - 2dak + 2dek - 2fck + 2fak = 0
Now get all the "k" terms to one side, and solve:
a²(e-c) + c²(a-e) + e²(c-a) + b²(e-c) + d²(a-e) + f²(c-a) = 2k(b(e-c)+d(a-e)+f(c-a))
(a²+b²)(e-c) + (c²+d²)(a-e) + (e²+f²)(c-a) = 2k(b(e-c)+d(a-e)+f(c-a))
k = (1/2)((a²+b²)(e-c) + (c²+d²)(a-e) + (e²+f²)(c-a)) /
(b(e-c)+d(a-e)+f(c-a))
Solve for h the same way:
h = (1/2)((a²+b²)(f-d) + (c²+d²)(b-f) + (e²+f²)(d-b)) /
(a(f-d)+c(b-f)+e(d-b))
Now, to find r², just plug your "h" and "k" into one of
the original equations:
r² = (a-h)² + (b-k)²
Measure of an Arc
You might wonder (or someone might ask you) what is the measure of each of
the three arcs cut off by the three points that determine the circle you just
found. The answer can be obtained without ever finding the circle
itself! How? If the three points A, B, C determine a circle, the
measure of arc AB is exactly twice the measure of angle ACB. That's just a
property of inscribed angles: The measure of an angle inscribed in a circle is half the measure of the arc it intercepts.
Related
pages in this website
Conic Sections -- the home page for
conic sections.
Circumscribed
Circle -- if the three points given here are the vertices of a triangle,
then the equation we find on this page is the equation of the circumscribed
circle.
Find the points
of intersection of two circles.
Inscribed
Angle Property -- The measure of an angle inscribed in a circle is half the measure of the arc it intercepts.
