# The logic problem faced in linear algebra

1. Sep 15, 2011

### peter13

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!

2. Sep 15, 2011

### peter13

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.