## Quadratic Circle

#### Problem

The quadratic curve, `y` = `x`^{2}`bx``c`, where `b`,`c` ^{⊕}, is drawn and the points of intersection with the axes are labelled, A, B, and C. A circle is drawn through the points A, B, and C, and a fourth point D is created on the y-axis.

Find the co-ordinate of D.

#### Solution

As the discriminant of the quadratic equation, `x`^{2}`bx``c`=0, is `b`^{2}+4`c`, which is positive, there will be two roots. The product of the roots, α_{1}α_{2}=-`c`, so one root will be negative and one will be positive. Let the three co-ordinates be A (α_{1},0), B (α_{2},0) and C (0,`c`). Hence the lengths, AO = -α_{1} (as the root will be negative), OB = α_{2}, and CO = c.

In a circle with two intersecting chords, the product of the segments are equal: AOOB = COOD, therefore, -α_{1}α_{2} = `c`OD.

As α_{1}α_{2}=-`c`, we get, OD = 1.

That is, the co-ordinate of D is (0,1), which, amazingly, is independent of both `b` and `c`.

What if `b` was allowed to be positive or negative?

What about `c`?

Investigate the general quadratic curve, `y` = `ax`^{2}+`bx`+`c`.

Prove the result that in a circle with two intersecting chords, the product of the segments are equal.