# Finding equation of a circle given three non - collinear points

A circle is given which passes through three non collinear points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ then prove that equation of this circle is given by $\begin{vmatrix} x^2+y^2&x&y&1\\ x_1^2+y_1^2&x_1&y_1&1\\ x_2^2+y_2^2&x_2&y_2&1\\ x_3^2+y_3^2&x_3&y_3&1\\ \end{vmatrix} = 0$ ?? It is just like one of answers to this question Finding an equation of circle which passes through three points I would like to know how we get to this determinant and I have no idea from where i can start??

## 1 Answer

Expanding the determinant in terms of $3\times 3$ determinants, using the top row, we obtain an equation $A(x^2+y^2)+B x +C y +D =0$ with constants $A,B,C,D.$ The solution set to an equation of this form is (i) empty,or (ii) contains one point, or (iii) is a line,or (iv) is a circle, or (v) is the whole plane. Since the top row co-incides with one of the other rows when $(x,y)=(x_i,y_i)$ for $i\in \{1,2,3\},$ the $3$ given points belong to the solution set, so we can rule out (i) and (ii). And (iii) or (v) each require $A=0,$ but the co-efficient $A\ne 0$ because the $3$ given points are not co-linear. This leaves (iv).And as already stated, the $3$ given points belong to the solution set, that is, they lie on the circle.

• why is solution set of $A(x^2+y^2)+Bx+Cy+Dz+E=0$ always of one of the five forms? also how did you rule out (i) and (ii) – Matt Apr 16 '16 at 18:16
• The determinant is $0$ when $(x.y)$ is one of the $3$ given points because then $2$ rows of the matrix are identical so each of the $3$ points belong to the solution set. This rules out (i),(ii), and actually rules out (iii) too because the $3$ points are not co-linear...... I had a typo. There is no "z"...... Complete the squares of the form "$0= A(x^2+....$ " when $A\ne 0$ to get $0=A(x-F)^2+A(y-G)^2+H$, which has unique solution $(F,G)$ when $H=0$, or is a circle when $A H<0$, and no solution when $A H>0.$ – DanielWainfleet Apr 17 '16 at 0:27
• No need for $E$ either ...... – DanielWainfleet Apr 17 '16 at 0:29