Center and Radius of a Circle

Question

How to find the center and radius of a circle through the points $(1,3)$, $(4,6)$ and $(-3,1)$? Could you give me some hint?

http://math.stackexchange.com/questions/213658/get-the-equation-of-a-circle-when-given-3-points — lab bhattacharjee, Dec 10 '13 at 08:47

score 1 · Answer 1 · answered Dec 10 '13 at 09:45

1

$$ \left( \begin{array}{ccc} 2x_1 & 2y_1 & -1 \\ 2x_2 & 2y_2 & -1\\ 2x_3 & 2y_3 & -1 \\ \end{array} \right) \left( \begin{array}{c} x \\ y \\ r^2 \\ \end{array} \right) = - \left( \begin{array}{c} x_1^2+y_1^2 \\ x_2^2+y_2^2 \\ x_3^2+y_3^2 \\ \end{array} \right) $$

answered Dec 10 '13 at 09:45

David Holden

18,040

the determinant of the matrix on the left is $-4(x_1y_2 -x_2y_1 +x_2y_3-x_3y_2 +x_3y_1-x_1y_3)$ which will be zero if two points coincide, or the three points are collinear – David Holden Dec 10 '13 at 09:53

score 0 · Answer 2 · answered Dec 10 '13 at 08:27

if you find the center the radius is easy.

geometrically the problem is simple to solve with ruler and compass, but turning that method into algebra is clumsy and error-prone.

a purer algebraic approach (making use of the symmetry in the situation) is to use the fact that the centre is equidistant from the three points: $$ (x-x_1)^2+(y-y_1)^2 = (x-x_2)^2+(y-y_2)^2=(x-x_3)^2 +(y-y_3)^2 $$

score 0 · Answer 3 · answered Dec 10 '13 at 08:29

Solve a system of equations: if the wanted circle is $\;(x-a)^2+(y-b)^2=r^2\;$ , then

$$\begin{align*}(1,3)\;\longrightarrow\;(1-a)^2+(3-b)^2=r^2\\ (4,6)\;\longrightarrow\;(4-a)^2+(6-b)^2=r^2\\ (-3,1)\,\longrightarrow\;(3+a)^2+(1-b)^2=r^2\end{align*}$$

Now equate, for example, equations (1)-(2):

$$a^2-2a+1+b^2-6b+9=a^2-8a+16+b^2-12b+36\implies$$

$$\implies 6(a+b)=42\implies a+b=7$$

Do now something similar with equations (1)-(3) or (2)-(3) and get another linear equation in two variables, and then solve...

Center and Radius of a Circle

3 Answers3