
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:
(xh)² + (yk)² = r²
Solve for h, k, and r in terms of a, b, c, d, e, and f.
Solution:
Start with these three equations:
(ah)² + (bk)² = r²
(ch)² + (dk)² = r²
(eh)² + (fk)² = r²
Multiply them out, and you get these three equations, equivalent to those above:
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 (ec), the second by (ae), and the third by (ca):
(a²  2ah + h² + b²  2bk + k²)(ec) = r²(ec)
(c²  2ch + h² + d²  2dk + k²)(ae) = r²(ae)
(e²  2eh + h² + f²  2fk + k²)(ca) = r²(ca)
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

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(ec) + c²(ae) + e²(ca) + b²(ec) + d²(ae) + f²(ca) = 2k(b(ec)+d(ae)+f(ca))
(a²+b²)(ec) + (c²+d²)(ae) + (e²+f²)(ca) = 2k(b(ec)+d(ae)+f(ca))
k = (1/2)((a²+b²)(ec) + (c²+d²)(ae) + (e²+f²)(ca)) / (b(ec)+d(ae)+f(ca))
Solve for h the same way:
h = (1/2)((a²+b²)(fd) + (c²+d²)(bf) + (e²+f²)(db)) / (a(fd)+c(bf)+e(db))
Now, to find r², just plug your "h" and "k" into one of the original equations:
r² = (ah)² + (bk)²
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.
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.
Find the equations of the (up to) four tangent lines to two circles.
Inscribed Angle Property  The measure of an angle inscribed in a circle is half the measure of the arc it intercepts.
The webmaster and author of this Math Help site is Graeme McRae.