Intersecting secants and coordinate geometry

The intersecting secants theorem states that when two secant lines intersect each other, the products of their segments are equal. In the picture below,

\[ PA \cdot PB = PC \cdot PD. \]

Secant lines AB and CD intersecting at a point P outside the circle.

The standard proof is as follows. Angles PAD and PCB are congruent because they are inscribed angles which cut off the same arc BD. The triangles PAD and PCB are similar because two pairs of corresponding angles are congruent. Therefore, PA/PD = PC/PB, hence \(PA \cdot PB = PC \cdot PD\).

We could also prove this result using coordinate geometry. Assume that P is at the origin, and suppose that the circle has equation \(x^2 + y^2 + ax + by + c = 0 \).

Any line through the origin can be parametrized as \(x = r\cos \theta\) and \(y = r\sin \theta\) where \(r\) is the distance from the origin and \(\theta\) is fixed.

Substituting into the equation of the circle yields
\[r^2 + ar \cos \theta + br \sin \theta + c = 0.\]

This is a quadratic equation in \(r\), and the product of the roots is \(c\). Therefore \(PA \cdot PB = PC \cdot PD = c\).