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Quadratic Circle

Problem

The quadratic curve, y = x2minusbxminusc, where b,c is in the set , 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, x2minusbxminusc=0, is b2+4c, which is positive, there will be two roots. The product of the roots, α1α2=-c, so one root will be negative and one will be positive. Let the three co-ordinates be A (α1,0), B (α2,0) and C (0,c). Hence the lengths, AO = -α1 (as the root will be negative), OB = α2, and CO = c.

In a circle with two intersecting chords, the product of the segments are equal: AOtimesOB = COtimesOD, therefore, -α1α2 = ctimesOD.

As α1α2=-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 = ax2+bx+c.
Prove the result that in a circle with two intersecting chords, the product of the segments are equal.

Problem ID: 168 (Apr 2004)     Difficulty: 4 Star

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