The radius of the circle is 8 cm. The distance of a point lying outside the circle from the centre is 17 cm. The length of the tangent drawn from the outside point to the circle is:

Correct — D. A tangent meets the radius at the point of contact at 90°, so the centre O, the outside point P and the contact point T form a right triangle in which OP = 17 cm is the hypotenuse.

OP² = OT² + PT²
17² = 8² + PT²
289 = 64 + PT²

PT² = 225
PT = 15 cm → option (d)

Why the others are wrong

• (a)
  16 cm — 16 cm is simply twice the 8 cm radius, lifted from the data rather than computed. Put it back and 8² + 16² = 320, not 289, so it fails Pythagoras.
• (b)
  19 cm — 19 cm is what √(17² + 8²) = √353 ≈ 18.8 rounds to — that is adding the squares. The radius is a leg, so the squares must be subtracted.
• (c)
  18 cm — 18 cm is longer than the 17 cm distance to the centre, and no leg can exceed the hypotenuse. PT² = 289 − 64 = 225, so PT = 15 cm.

The whole configuration collapses into one right triangle.

A tangent is perpendicular to the radius at the point of contact. That gives a right angle at T, with the radius OT = 8 cm as one leg, the tangent PT as the other, and the distance OP = 17 cm as the hypotenuse.

So the tangent length from an external point is always √(d² − r²), where d is the distance from the point to the centre.

Here 17² − 8² = 289 − 64 = 225, and 8, 15, 17 is a Pythagorean triple, so the root is the exact integer 15.

Key facts

• A tangent is perpendicular to the radius drawn to the point of contact.
• Tangent length from an external point = √(d² − r²), with d the distance to the centre.
• 8, 15, 17 is a Pythagorean triple, so √(17² − 8²) is exactly 15.
• The tangent length is always shorter than the distance from the point to the centre.

Common traps

• Adding the squares instead of subtracting, which gives about 18.8 cm.
• Treating the 8 cm radius as the hypotenuse of the triangle.
• Accepting an answer longer than the 17 cm distance to the centre.

No directly related past PYQ was found.