A point P is 13 cm away from the center of a circle. A tangent is drawn from P to the circle, and its length is 12 cm. What is the area of the circle?

Correct — A. Let O be the centre and T the point of contact. The radius OT meets the tangent at 90°, so OP is the hypotenuse of right triangle OTP.

Pythagoras: OT² + PT² = OP²
Substitute: r² + 12² = 13²
Subtract: r² = 169 − 144 = 25

Area = π × r² = π × 25 (no need to find r = 5)
= 25π cm² → option (a)

Why the others are wrong

• (b)
  144π cm² — 144π squares the tangent, treating 12 cm as the radius. The tangent runs from P to the point of contact; it is a leg of the triangle, not a radius.
• (c)
  36π cm² — 36π needs a radius of 6 cm, half the tangent. The right angle at the point of contact fixes r² = 169 − 144 = 25, not 36.
• (d)
  169π cm² — 169π takes OP = 13 cm as the radius. P lies outside the circle, so OP is the hypotenuse and is longer than the radius.

A tangent is perpendicular to the radius at the point of contact. So for any external point P, the tangent length t, the radius r and the distance d from the centre satisfy d² = r² + t².

Here d = 13 and t = 12, two members of the 5–12–13 Pythagorean triple, so r = 5 cm and the area is 25π cm².

Key facts

• The radius drawn to the point of contact is perpendicular to the tangent.
• Tangent length from an external point: t² = d² − r², where d is the distance from the centre.
• 5, 12, 13 is a Pythagorean triple: 25 + 144 = 169.
• Area of a circle = πr², so the working can stop at r².

Common traps

• Taking the distance OP as the radius
• Adding the squares (169 + 144) as if OP were a leg, when it is the hypotenuse

No directly related past PYQ was found.