Two concentric circles with a common centre 'O' have a chord AB on the outer circle, which intersects the inner circle at points C and D. If the distance of the chord from the centre is 3 cm, the outer radius is 13 cm and the inner radius is 7 cm, then the length (in cm) of AC is:
$2\sqrt10$
Two circles share the same centre O (concentric circles). A straight line cuts the outer circle at A and B, and along the way cuts the inner circle at C and D. So along the line the points occur in the order A, C, D, B. We are told the perpendicular distance of this line from O is 3 cm, the outer radius is 13 cm and the inner radius is 7 cm.
Key theorem: the perpendicular drawn from the centre of a circle to a chord bisects that chord. Let M be the foot of the perpendicular from O to the line; then OM = 3 cm, and M is the common midpoint of both chords AB and CD.
Step 1 — Half of the outer chord (AM). Triangle OMA is right-angled at M, with hypotenuse OA = 13 (outer radius):
AM = √(OA² − OM²) = √(13² − 3²) = √(169 − 9) = √160 = 4√10 cm.
Step 2 — Half of the inner chord (CM). Triangle OMC is right-angled at M, with hypotenuse OC = 7 (inner radius):
CM = √(OC² − OM²) = √(7² − 3²) = √(49 − 9) = √40 = 2√10 cm.
Step 3 — Length AC. Since A, C and M lie on the line with C between A and M:
AC = AM − CM = 4√10 − 2√10 = 2√10 cm (≈ 6.32 cm).
The correct answer is 2√10 cm. A common mistake is to add the two halves (4√10 + 2√10) — that would give the length AD, not AC — or to forget that OM = 3 must be subtracted inside the square root. By symmetry BD also equals 2√10, since the perpendicular from the centre makes both chords symmetric about M.
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