The graph of a(x^2+y^2)+bx+cy+d=0 is a circle if a not=0. Show that there is a c

Question

The graph of a(x^2+y^2)+bx+cy+d=0 is a circle if a not=0. Show that there is a circle through any three points in the plane that are not all on a line.

Explanation / Answer

Lets consider the following, we wish to find a quadratic : (x^2+y^2)+bx+cy+d=0 passing through the points (x1,y1) (x2,y2) and (x3,y3). First lets settle existence. As has been pointed out above, if the points are not colinear we can use a circle. If the points are colinear then consider one line through the 3 points and some other line. The union of the two is a quadratic (union = multiplying the equations), so it will do. Next question, how many such quadratics are there? Plugging into our three points into (#) we obtain a system of 3 equations in 6 unkowns. The coefficents are the unknowns here. We have too many unkowns and too few equations. More precisely, the solutions are the kernel of a linear transform R^6 -> R^3 so it has dimension at least three. The upshot, our system has infinitely many solutions. To answer the question, we need to observe that we excluded the possibility of the only solutions being of the form a=b=c=0 in the first paragraph. A little more linear algebra shows that in fact there are infinitely many solutions not of the form a=b=c=0. What about 6 points? Probably the answer can be found in some text on classical algebraic geometry. There will be some need to be some condition on the points, I have no idea what it is. For example if three are colinear and the other three are random then it is not possible

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