Multiple choice

In figure there are two concentric circles with centre O and of radii 5 cm and 3 cm. From an external point P tangents PA and PB are drawn to these circles. If AP = 12 cm find the length of BP

  1. $\displaystyle 4\sqrt{10}$ cm
  2. $\displaystyle 2\sqrt{10}$ cm
  3. $\displaystyle \sqrt{10}$ cm
  4. $\displaystyle 3\sqrt{10}$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the outer circle, radius R=5, tangent AP=12. By Pythagoras, OP^2 = 5^2 + 12^2 = 25 + 144 = 169, so OP = 13. In the inner circle, radius r=3, tangent BP. By Pythagoras, OP^2 = r^2 + BP^2. 169 = 3^2 + BP^2. 169 = 9 + BP^2. BP^2 = 160. BP = sqrt(160) = 4 * sqrt(10).

AI explanation

The tangent and the radius at the point of contact form a right-angled triangle with the line from the external point to the center. Using the Pythagorean theorem for the outer circle, OP squared equals OA squared plus AP squared, so OP squared is 5 squared plus 12 squared, which gives an OP squared value of 169. For the inner circle with radius 3 cm, BP squared equals OP squared minus OB squared, becoming 169 minus 9 to equal 160, and the square root of 160 simplifies to 4 sqrt(10) cm.

Keep practicing — related questions