Navigation 
 Home 
 Search 
 Site map 

 Contact Graeme 
 Home 
 Email 
 Twitter

 Skip Navigation LinksMath Help > Geometry > Circles, Conic Sections > Eq of Circle given 3 points

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 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 (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
-------------------------------------------------------------------------------------------------------------------------------------------
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.

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.