If two secants from an external point P intersect a circle at A,B and C,D respectively, which relation holds?

• A. PA × PC = PB × PD
• B. PA + PB = PC + PD
• C. PA × PD = PB × PC
• D. PA × PB = PC × PD

Answer: (D)

The key idea here is the power of a point with respect to a circle. If a point P lies outside the circle and you draw two secants through P meeting the circle at A and B on one line and at C and D on the other, the product of the distances from P to the nearer and farther intersection on each secant is the same: PA × PB = PC × PD. A quick way to see this is to draw a tangent from P to the circle, touching at T. The tangent-secant theorem gives PT^2 = PA × PB and also PT^2 = PC × PD, so the two products are equal. Therefore the relation that holds is PA × PB = PC × PD.