The Tangent That Divides
A circle with centre O and radius 5 is given. Point P is outside the circle with OP = 13. From P, a tangent touches the circle at T, and a secant through P meets the circle at A (nearer to P) and B (farther). The length PA is 4. A student computes PT using the tangent–secant theorem and then tries to find the angle ∠TPO. A second part of the problem asks for the area of triangle POB and whether the secant can be rotated so that PA = PB.
1. Find PT and verify it using the right triangle OPT. Then find AB.
Read the worked answer
Tangent–secant: PT² = PA · PB. First find PB. Since PA = 4 and AB is unknown, write PB = 4 + AB.
From right triangle OPT: OT ⊥ PT, so PT² = OP² − OT² = 169 − 25 = 144, PT = 12.
Then 144 = 4 · PB ⇒ PB = 36 ⇒ AB = 32.
Check: the chord AB has length 32, but the diameter is only 10, so a chord of length 32 is impossible. The data are inconsistent: a secant from P with PA = 4 cannot have PB = 36 for this circle. The maximum chord through the circle from P is bounded, so the given PA = 4 is incompatible with OP = 13 and r = 5; the tangent length PT = 12 is correct, but the secant data must be revised.
2. Find the angle ∠TPO and the area of triangle POB using the correct tangent configuration.
Read the worked answer
In right triangle OPT, sin∠TPO = OT/OP = 5/13, so ∠TPO = arcsin(5/13) ≈ 22.6°.
For triangle POB, OB = 5 and OP = 13. The angle at O between OB and OP depends on where B lies. If B is the far intersection along a secant, the maximum possible PB occurs when the secant passes through O, giving PB = PO + OB = 18 and PA = PO − OB = 8. Thus PA cannot be 4; the smallest possible PA along a secant through the circle is 8.
Taking the diameter secant, triangle POB has sides PO = 13, OB = 5, PB = 18, which is degenerate (13 + 5 = 18), area 0. For a genuine secant, the area varies; the configuration with PA = 4 is impossible.