Multiple choice

Two circles touch externally. The sum of their area is $130\;\pi$ sq. cm and the distance between their centres is $14$ cm. The radii of the circles are

  1. $11$ cm and $3$ cm
  2. $10$ cm and $2$ cm
  3. $9$ cm and $1$ cm
  4. $8$ cm and $1$ cm
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A Correct answer
Explanation

Let radii be r1 and r2. r1 + r2 = 14. Area sum = pi(r1^2 + r2^2) = 130pi, so r1^2 + r2^2 = 130. (r1+r2)^2 = r1^2 + r2^2 + 2r1r2. 196 = 130 + 2r1r2, so 2r1r2 = 66, r1r2 = 33. Solving r^2 - 14r + 33 = 0 gives (r-11)(r-3)=0. Radii are 11 and 3.

AI explanation

Let the radii of the two circles be r and R. Since they touch externally, r + R = 14. The sum of their areas is pi(r^2 + R^2) = 130 pi, so r^2 + R^2 = 130. Using the identity (r + R)^2 = r^2 + R^2 + 2rR, we have 14^2 = 130 + 2rR, yielding 2rR = 66. The radii are the roots of x^2 - 14x + 33 = 0, which factors to (x - 11)(x - 3) = 0. The radii are 11 cm and 3 cm.

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