The angle between two tangents drawn from an external point to a circle is 60°. What is the angle subtended by the chord connecting their points of contact at the center?

Correct — B. Let the tangents from P touch the circle at A and B, and let O be the centre.

Radius ⊥ tangent: ∠OAP = ∠OBP = 90°
Angle sum of quadrilateral OAPB = 360°
∠AOB = 360° − 90° − 90° − 60°
= 120° → option (b)

Why the others are wrong

• (a)
  60° — 60° is the angle at P itself. The angle at the centre is its supplement: the two right angles leave 180° − 60° = 120°.
• (c)
  90° — 90° at the centre would need the tangents to meet at 90° too, making OAPB a square. Here they meet at 60°, so ∠AOB = 180° − 60° = 120°.
• (d)
  30° — 30° is half the angle at P, which is ∠APO, since OP bisects ∠APB. It is not the angle the chord subtends at the centre.

The radius to a point of contact is perpendicular to the tangent. So the quadrilateral formed by the centre, the two points of contact and the external point has two right angles, and its remaining two angles add to 180°.

That gives the working rule ∠AOB = 180° − ∠APB: the angle between the tangents and the angle the chord of contact subtends at the centre are supplementary.

Key facts

• ∠AOB + ∠APB = 180° for tangents PA and PB from an external point P.
• OP bisects both ∠APB and ∠AOB.
• Tangents from one external point are equal: PA = PB.
• When ∠APB = 60°, triangle PAB is equilateral.

Common traps

• Giving the angle at P, 60°, as the angle at the centre
• Halving 60° to 30°, which is ∠APO, not ∠AOB

No directly related past PYQ was found.