339. The Tangent-Secant Secret · PA × PB always equals PT²

Try to break it

Drag P around outside the circle. The tangent length PT and the secant lengths PA, PB all change, but the product PA × PB always equals PT². Try to find a position where the equality breaks; impossible. (Push the secant to a tangent and A coincides with B — both equal T — so PA × PB = PT² becomes PT × PT = PT² trivially.)

How you build it

Draw a tangent and a secant from an external point P, then check that PT² = PA × PB.

• Point tool: mark O near the middle — the centre of the circle.
• Circle tool: click O as the centre, then click outward to set the radius.
• Point tool: mark P well outside the circle — the external point.
• Line tool: from P, draw a line that just grazes the circle — touching it at a single point.
• Point tool: mark T where the tangent touches the circle. PT is the tangent length.
• Line tool: from P, draw a second line that cuts straight through the circle, crossing it at two points.
• Point tool: mark A where the secant first meets the circle — the crossing nearer to P.
• Point tool: mark B where the secant leaves the circle (the far crossing). Now PA × PB = PT² — the tangent squared equals the product of the secant's two segments.

The proof, step by step

Prove that the square of the tangent equals the product of the secant segments.

• Join TA and TB to form triangles PAT and PTB.
• By the alternate segment theorem, ∠PTB = ∠TAB (angles in alternate segment).
• ∠P is common to both triangles.
• Therefore, ΔPAT ~ ΔPTB by AA similarity postulate.
• Corresponding sides are proportional: PA/PT = PT/PB.
• Cross-multiplying gives PA × PB = PT². Hence proved.

Worked example

A tangent PT and a secant PAB intersect at point P outside a circle. If PA = 4 cm and PB = 9 cm, what is the length of PT?

Using the Tangent-Secant Theorem: PA × PB = PT². Substituting values: 4 × 9 = PT² → PT² = 36 → PT = 6 cm.

• 6 cm — correct
• 13 cm
• 36 cm
• 5 cm