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Two circles share a centre. A chord of the big one can touch the small one.
- Two circles, same centre O — outer radius 17, inner radius 8.
- Chord AB of the outer circle touches the inner circle at M.
- OM meets the chord at 90°, and M is the chord's midpoint.
- MA² = 17² − 8² = 225, MA = 15. Chord AB = 30.
One line, two jobs
Two circles with the same centre are concentric. Let the outer chord AB touch the inner circle at M. Then AB is a tangent to the inner circle.
The tangent AB meets the radius OM at 90°. So the angle at M is a right angle. A perpendicular from the centre halves a chord. So M is the midpoint of AB. One line, OM, does both jobs at once.
Triangle OMA is right-angled at M. OA is the outer radius, the hypotenuse. OM is the inner radius. MA is half the chord — found like a tangent length.
Worked through
Outer radius OA = 17. Inner radius OM = 8. By Pythagoras:
MA² = OA² − OM² = 17² − 8² = 289 − 64 = 225, so MA = 15.
The chord is twice the half: AB = 2 × 15 = 30.
Slide both radii. Read the chord.
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