Linear Algebra and the C Language/a06a

__NOTOC__ The coefficients of the equation of a circle. Presentation : A homogeneous linear system with as many equations as unknowns has a nontrivial solution if and only if the determinant of the matrix is zero. Let us calculate the equation of the circle passing through points P, Q and R : c1(x^2 +y^2) + c2 x + c3 y + c4 = 0 This same equation with the points P(x1,y1) Q(x2,y2) and R(x3,y3): c1(x1^2+y1^2) + c2 x1 + c3 y1 + c4 = 0 c1(x2^2+y2^2) + c2 x2 + c3 y2 + c4 = 0 c1(x3^2+y3^2) + c2 x3 + c3 y3 + c4 = 0 The system of four equations: c1(x^2 +y^2) + c2 x + c3 y + c4 = 0 c1(x1^2+y1^2) + c2 x1 + c3 y1 + c4 = 0 c1(x2^2+y2^2) + c2 x2 + c3 y2 + c4 = 0 c1(x3^2+y3^2) + c2 x3 + c3 y3 + c4 = 0 The determinant of the system: |x^2 + y^2 x y 1| |x1^2+y1^2 x1 y1 1| = 0 |x2^2+y2^2 x2 y2 1| |x3^2+y3^2 x3 y3 1| The determinant in C language: | 1 1 1 1| |x1^2+y1^2 x1 y1 1| = 0 |x2^2+y2^2 x2 y2 1| |x3^2+y3^2 x3 y3 1| To calculate the coefficients of the equation of the circle, we use the cofactor expansion along the first row. cof(R1,C1)(x^2 + y^2) + cof(R1,C2) x + cof(R1,C3) y + cof(R1,C4) = 0 This equation gives us the equation of the circle that passes through the three points P, Q and R. Application dcircle.h .............. Temporary h file Delete this file at the end of the session. c00a.c c00b.c