Question:
The line AB is 6 metres and is tangent to the inner of two concentric circles at point C. Radii of the two circles are integers. What is the radius of the outer circle?
The line AB is 6 metres and is tangent to the inner of two concentric circles at point C. Radii of the two circles are integers. What is the radius of the outer circle?
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Use Pythagoras on concentric circle tangents, then apply difference of squares to solve radius relations.
Updated On: Aug 6, 2025
- 5 metres
- 4 metres
- 6 metres
- 3 metres
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The Correct Option is A
Solution and Explanation
Let:
- \( r \) be the radius of the inner circle,
- \( R \) be the radius of the outer circle,
- Line AB = 6 m,
- Point C is the point of tangency from AB to the inner circle.
Since both circles are concentric, and AB is a tangent to the inner circle, triangle OCB is a right triangle where: - OB = R (outer radius), - OC = r (inner radius), - CB = half of AB = 3 (since AB is tangent to both sides symmetrically). Using Pythagoras in triangle OCB: \[ OB^2 = OC^2 + CB^2 R^2 = r^2 + 3^2 = r^2 + 9 R^2 - r^2 = 9 (R - r)(R + r) = 9 \] Now factor 9 as a product of integers: \[ (1,9) R - r = 1, R + r = 9 R = 5, r = 4 \] \[ (3,3) R - r = 3, R + r = 3 R = 3, r = 0 \text{ (not valid)} \] Only valid integer solution: \[ \boxed{R = 5 \text{ metres}} \]
Since both circles are concentric, and AB is a tangent to the inner circle, triangle OCB is a right triangle where: - OB = R (outer radius), - OC = r (inner radius), - CB = half of AB = 3 (since AB is tangent to both sides symmetrically). Using Pythagoras in triangle OCB: \[ OB^2 = OC^2 + CB^2 R^2 = r^2 + 3^2 = r^2 + 9 R^2 - r^2 = 9 (R - r)(R + r) = 9 \] Now factor 9 as a product of integers: \[ (1,9) R - r = 1, R + r = 9 R = 5, r = 4 \] \[ (3,3) R - r = 3, R + r = 3 R = 3, r = 0 \text{ (not valid)} \] Only valid integer solution: \[ \boxed{R = 5 \text{ metres}} \]
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