Multiple choice

Let C be a circle centre O. Let T be a point on the circle and P a point outside the circle such that PT is tangent to C. Assume that the segment OP intersects C in a point Q If PT = 12 and PQ = 8 the radius of C is

  1. $r = 40$
  2. $r = 5$
  3. $\displaystyle r=4\sqrt{5}$
  4. $\displaystyle r=4\sqrt{13}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By the power of a point theorem, PT^2 = PQ * (PQ + 2r). Given PT = 12 and PQ = 8, 144 = 8 * (8 + 2r). 18 = 8 + 2r, so 10 = 2r, r = 5.

AI explanation

Let the radius of the circle be r. The length of the tangent from external point P is PT = 12, and since the radius OT is perpendicular to tangent PT, triangle OPT is a right triangle. Using the Pythagorean theorem, OP squared equals OT squared plus PT squared, so (PQ + OQ) squared equals r squared plus 12 squared. Substituting the given values gives (8 + r) squared equals r squared plus 144, which expands to 64 plus 16r plus r squared equals r squared plus 144. Solving 16r equals 80 gives the radius as r = 5.

Keep practicing — related questions