I have some difficulties to understand the logic of the solution to this problem.(adsbygoogle = window.adsbygoogle || []).push({});

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!

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# The logic problem faced in linear algebra

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