R is a point outside a circle and is 18 cm away from its centre. A secant drawn from the point R intersects the circle at points A and B in such a way that RA = 7 cm and AB = 5 cm. The radius of the circle (in cm) is:

Correct — D. Use the power of the external point R: RA × RB = OR² − r², where B is the far intersection.

RA = 7 and AB = 5, so RB = RA + AB = 12 cm
Power of R = 7 × 12 = 84

Power also equals OR² − r² = 18² − r² = 324 − r²

324 − r² = 84
r² = 324 − 84 = 240
r = √240 = √(16 × 15) = 4√15 cm, the surd shown in option (d)

Why the others are wrong

• (a)
  Option (a) is 5√15, so r² = 375 — larger than OR² = 324. That would place R inside the circle, contradicting a point stated to be outside it.
• (b)
  Option (b) is 7√15, giving r² = 735, more than twice OR². The same objection is fatal: the radius cannot exceed the distance from the centre to an external point.
• (c)
  Option (c) is 3√15, so r² = 135 and the power of R would be 324 − 135 = 189. The secant fixes that power at 84, so 135 is not the radius squared.

For a point R outside a circle, every line through R that cuts the circle satisfies RA × RB = a constant, with A the nearer intersection and B the farther. That constant is the power of the point, and it equals OR² − r².

The trap is RB. You are given RA and the chord AB, not RB, so RB = RA + AB = 7 + 5 = 12. Multiplying 7 by 5 is the commonest error on this item.

The tangent form of the same fact is RT² = OR² − r², which is why a tangent length and a secant product are interchangeable in these questions.

Key facts

• Power of an external point: RA × RB = OR² − r², with RB measured to the far intersection.
• RB = RA + AB, so 7 + 5 = 12 and the power of R is 84.
• 324 − 84 = 240, and √240 simplifies to 4√15.

Common traps

• Using AB in place of RB and computing 7 × 5 = 35.
• Reading the 18 cm as the radius rather than as the distance from the centre.
• Stopping at √240 without reducing it to 4√15.

No directly related past PYQ was found.