A line is drawn from a point P located 13 cm away from the center O of a circle with radius 5 cm. If the line is tangent to the circle, what is the length of the tangent from point P to the point of contact with the circle?

Correct — C. Call the point of contact T. The radius OT is perpendicular to the tangent, so triangle OTP is right-angled at T.

Hypotenuse: OP = 13 cm, leg OT = 5 cm
Pythagoras: PT² = 13² − 5² = 169 − 25 = 144
Root: PT = √144 = 12 cm → option (c)

Why the others are wrong

• (a)
  8 cm — 8 cm is 13 − 5, lengths subtracted before squaring. Pythagoras subtracts the squares, 169 − 25 = 144, so the tangent is 12.
• (b)
  9 cm — 9 cm fails the right-angle check. With the 5 cm radius it gives 9² + 5² = 81 + 25 = 106, not 13² = 169.
• (d)
  10 cm — 10 cm fails the right-angle check. 10² + 5² = 100 + 25 = 125, not 169. The tangent 12 passes it: 144 + 25 = 169.

A tangent touches the circle at one point, and the radius to that point is perpendicular to the tangent.

So the centre O, the outside point P and the point of contact T form a right triangle: OP is the hypotenuse, OT = r and PT is the tangent length.

Tangent length = √(OP² − r²), and the two tangents from P are equal.

Key facts

• The radius at the point of contact is perpendicular to the tangent.
• Tangent length from an outside point P = √(OP² − r²).
• The two tangents drawn from one outside point are equal in length.
• 5-12-13 is a Pythagorean triple: 25 + 144 = 169.

Common traps

• Subtracting 13 − 5 = 8 without squaring.
• Adding the squares: √(13² + 5²) = √194 is longer than OP, and a tangent segment PT is always shorter than OP.

No directly related past PYQ was found.