I have some difficulties to understand the logic of the solution to this problem. Q:The graph of a(x2 +y2)+bx+cy +d = 0 is a circle if a 6= 0. Show that there is a circle through any three points in the plane that are not all on a line. solution:Insisting that the graph of a(x2+y2)+bx+cy+d = 0 passes through the three points gives a homogeneous linear system of three linear equations in four variables. This has a nontrivial solution, and a = 0 would be the three points all lie on the line bx+cy +d = 0, contrary to the hypothesis. So a = 0, as desired. if there is some problem of the solution ,pls point it out and explain to me in a detail way. Thanks a lot and this is the first time i use this forum!
sorry for that some typing mistake. the solution shuld be:solution:Insisting that the graph of a(x2+y2)+bx+cy+d = 0 passes through the three points gives a homogeneous linear system of three linear equations in four variables. This has a nontrivial solution, and a = 0 would be the three points all lie on the line bx+cy +d = 0, contrary to the hypothesis. So a is not equal to 0, as desired.