In the adjoining figure, the sum of radii of two concentric circles is 16 cm. The length of chord AB which touches the inner circle at P is 16 cm. The difference of the radii of the given circles is
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
Let R and r be radii of outer and inner circles. Given: R + r = 16, AB = 16, so AP = PB = 8 cm.
Since OP ⊥ AB (radius ⊥ tangent), in right △OPB: $R^2 = r^2 + 8^2$
$(R-r)(R+r) = 64$ → $(R-r) \times 16 = 64$ → $R - r = 4$ cm.
Answer: B) 4 cm
Explanation
- Since AB touches the inner circle at P, OP ⊥ AB (Theorem 10.1), so P is the midpoint of AB, giving BP = 8 cm.
- Apply Pythagoras in △OPB: $R^2 - r^2 = 64$, i.e., $(R+r)(R-r) = 64$.
- Substitute R + r = 16 to get R − r = 4 cm. This is the key identity to remember for concentric circles with tangent-chord problems.