From a point P outside a circle, a tangent PA and a secant PBC are drawn. If PA=6 cm and BC=5 cm, what is the length of the segment PB?

Correct — A. Tangent–secant theorem: PA² = PB × PC, the tangent squared equals the outside part times the whole secant.

PC = PB + BC. Let PB = x, so PC = x + 5
Substitute: 6² = x(x + 5)
Expand: x² + 5x − 36 = 0

Factor: (x + 9)(x − 4) = 0
Reject x = −9 (a length is positive), so x = 4
Check: PB × PC = 4 × 9 = 36 = 6² ✓

PB = 4 cm → option (a)

Why the others are wrong

• (b)
  5 cm — 5 cm is BC, the chord inside the circle, not PB. Check: PB = 5 gives PB × PC = 5 × 10 = 50, not 36.
• (c)
  9 cm — 9 cm is PC, the whole secant (4 + 5), and the size of the rejected root x = −9. PB is only the part outside the circle.
• (d)
  12 cm — 12 cm fails the theorem: 12 × (12 + 5) = 204, not 36. PB must be shorter than PA, because PB² is less than PB × PC = PA².

From an external point P, tangent² = external part × whole secant: PA² = PB × PC. The two-secant rule, PB × PC = PD × PE, becomes this one when the second secant's two meeting points merge into the point of contact.

When the chord BC is given rather than PC, the unknown appears in both factors and the equation turns quadratic.

Key facts

• Tangent–secant theorem: PA² = PB × PC, with B the nearer intersection point.
• Two secants PAB and PCD from one point: PA × PB = PC × PD.
• The outside part of a secant is shorter than the tangent from the same point.

Common traps

• Using PA² = PB × BC, the outside part times the chord, which gives PB = 7.2 cm, not an option
• Picking 9 cm, which is the whole secant PC, not PB

No directly related past PYQ was found.