From a point outside a circle, two tangents are drawn to the circle. If one of the tangents measures 12 cm, what is the length of the other tangent?

Correct — C. Call the external point P, the contact points A and B, and the centre O, then join OA, OB and OP.

Radii: OA = OB
Tangent meets radius at 90°: ∠OAP = ∠OBP = 90°
Shared hypotenuse: OP
So △OAP ≅ △OBP by RHS

Corresponding sides: PB = PA = 12 cm → option (c)

Why the others are wrong

• (a)
  10 cm — 10 cm would make the two tangents unequal, but △OAP ≅ △OBP by RHS forces PA = PB. The other tangent must match the given 12 cm.
• (b)
  11 cm — 11 cm breaks the equal-tangents theorem by 1 cm. Both tangents are legs of congruent right triangles sharing the hypotenuse OP, so both are 12 cm.
• (d)
  14 cm — 14 cm is impossible with the same radius and the same OP: Pythagoras gives each tangent √(OP² − r²), one value, so both tangents are 12 cm.

From a point outside a circle, two tangents can be drawn, and the lengths from that point to the two contact points are equal.

The proof is one congruence. The two radii are equal, each meets its tangent at 90°, and OP is common, so the two right triangles are congruent by RHS. The same congruence shows that OP bisects the angle between the tangents.

Key facts

• Tangents drawn from an external point to a circle are equal in length.
• The line joining the centre to the external point bisects the angle between the two tangents.
• The angle between the two tangents and the angle between the two radii at the centre add up to 180°.
• Each tangent length = √(OP² − r²).

Common traps

• Hunting for a radius or a distance to calculate with, when the equal-tangents theorem settles it in one line.
• Treating the tangent length as the distance from P to the centre: the tangent ends at the contact point, and OP is longer.

No directly related past PYQ was found.