3
$\begingroup$

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?

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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) $\endgroup$
    – Matt
    Commented Apr 16, 2016 at 18:16
  • $\begingroup$ 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 $A H>0.$ $\endgroup$ Commented Apr 17, 2016 at 0:27
  • $\begingroup$ No need for $E$ either ...... $\endgroup$ Commented Apr 17, 2016 at 0:29
1
$\begingroup$

The quadratic form $Ax^2+By^2+Cxy+Dx+Ey+F=0$ represents always a conic (eventualy degenerate in two right lines if it is a product of two linear equations) and it represents a circle if and only if $A=B\ne0$ and $C=0$. Then for a circle we have the equation $$A(x^2+y^2)+Dx+Ey+F=0$$ When we want a circle passing through three non collinear points $(x_i,y_i)$ we have a linear system of three equations whose solution gives the three values $ a=\dfrac{D}{A}, b=\dfrac{E}{A}, c=\dfrac{F}{A}$ of the required circle.

But when we want to have the equation of the circle without getting explicitely $a,b,c$ we consider the linear system of four equations with three unknowns $a,b,c$ $$(x^2+y^2)+ax+by+c=0\\(x_1^2+y_1^2)+ax_1+by_1+c=0\\(x_2^2+y_2^2)+ax_2+by_2+c=0\\(x_3^2+y_3^2)+ax_3+by_3+c=0$$ and it is well known that the condition of compatibility of this four equations is that its determinant be equal to zero. In other words, making more explicitely, we have the matricial equation

$$\begin{pmatrix} x^2+y^2&x&y&1\\ x_1^2+y_1^2&x_1&y_1&1\\ x_2^2+y_2^2&x_2&y_2&1\\ x_3^2+y_3^2&x_3&y_3&1\\ \end{pmatrix}\cdot\begin{pmatrix}1\\a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}$$

There are two evident solutions, $(0,0,0,0)$ and $(1,a,b,c)$ so the determinant must be equal to $0$ because if not then the solution should be unique.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .