Two circles intersect at two points, one of them X. Find Y on one circle and Z on the other, so that X, Y and Z are collinear and XY.XZ is as large as possible.
Solution
Let the circle through XY have center O and the other circle have center O'. XY = 2 OX sin XOY, XZ = 2 O'X sin XOZ, so we wish to maximise 2 sin XOY sin XO'Z. But 1/2 ∠XOY = 90o - ∠OXY, 1/2 ∠XO'Z = 90o - ∠OXZ, so 1/2 ∠XOY + 1/2 ∠XO'Z = ∠OXO', which is fixed. Thus ∠XOY + ∠XO'Z is fixed. But 2 sin XOY sin XO'Z = cos(XOY-XO'Z) - cos(XOY+XO'Z), so we maximise by taking XOY = XO'Z.
Note that in this case ∠OYZ = ∠O'ZY, so if we extend YO and ZO' to meet at C, then CY = CZ and hence C is the center of a circle containing the two circles and touching them at Y and Z. Also ∠CZY = ∠OXY, so O'Z is parallel to OX. Similarly OY is parallel to O'X, which shows how to construct Y and Z.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002