Some parabolas of the form y=x
2+bx+c have three distinct intercepts (two x and one y). If for all such parabolas a circle is drawn through these three points, they will have a point in common.
Prove this assertion and find the point.
My exploratory thoughts were to reduce b and c to zero, then the parabola becomes x^2 and its associated circle becomes a degenerate point at the origin, which is the vertex of y=x^2.
Then the assertion is the vertex of the parabola is always within the circle containing the intercepts.
I found it easier to work with the parabola in a factored form: y=(x-u)(x-v)=x^2-(u+v)x+uv.
Then the intercepts are (u,0) (v,0) and (0,uv).
The vertical line x=(u+v)/2 will have the center on it. Let w be that y value.
Then the distances from ((u+v)/2,w) to either (u,0) or (0,uv) are the same. So by the distance formula we can create:
sqrt[((u-v)/2)^2+w^2] = sqrt[((u+v)/2)^2+(w-uv)^2]
Solving for w we get w=uv/2+1.
Plugging w=uv/2+1 back into one of the two expressions for distance yields a radius of sqrt[u^2/4-uv/2+v^2/4+(uv)^2/4+1].
The distance from the vertex to the center is just w=uv/2+1. So then for the assertion to be true we must have:
uv/2+1 <= sqrt[u^2/4-uv/2+v^2/4+(uv)^2/4+1]
Square each side and move everything to one side:
0 =< u^2/4+uv/2+v^2/4 = (u/2+v/2)^2
The right side is always nonnegative, so the inequality is always true, and therefore the assertion that the vertex is always within the circle is true.
Solution
