The radii of two concentric circles are 26 cm and 10 cm. If the chord of the greater circle is a tangent to the smaller circle, then find the length of that chord.
- (a)48 cm
- (b)28 cm
- (c)38.7 cm
- (d)28.7 cm
Answer
Why
Correct — A. A tangent meets the radius at the point of contact at 90°, and a perpendicular from the centre bisects a chord.
Both circles share centre O, so the small radius OM drawn to the point of contact is that perpendicular, and M is the midpoint of chord AB.
Right triangle OMA: OA = 26 cm, OM = 10 cm
AM² = 26² − 10² = 676 − 100 = 576, so AM = 24 cm
Chord AB = 2 × 24 = 48 cm → option (a).
Why the others are wrong
- (b)28 cm — 28 makes the half-chord 14, and 14² + 10² = 296. The hypotenuse OA demands 26² = 676, so 14 is far too short.
- (c)38.7 cm — 38.7 halves to about 19.35, and 19.35² + 10² ≈ 474, not 676. The decimal also misses that 676 − 100 = 576 is an exact square.
- (d)28.7 cm — 28.7 halves to about 14.35, giving 14.35² + 10² ≈ 306. That would put the big radius near 17.5, not the 26 the question states.
Concept
Two theorems do all the work here. A tangent is perpendicular to the radius drawn to the point of contact, and the perpendicular from the centre bisects the chord.
Concentric circles let the two theorems meet. The small radius reaching the point of tangency is perpendicular to the chord, and because that radius starts at the shared centre it also bisects the chord.
What remains is a right triangle with the big radius as hypotenuse and the small radius as one leg, so the half-chord is √(R² − r²).
The answer only needs R² − r², never R and r separately, which is why the numbers 26 and 10 were chosen: 676 − 100 = 576 = 24².
Key facts
- For concentric circles of radii R and r, a chord of the larger circle that touches the smaller has length 2√(R² − r²).
- A tangent is perpendicular to the radius drawn to the point of contact.
- The perpendicular from the centre of a circle to a chord bisects that chord.
- Here 2√(26² − 10²) = 2√576 = 48 cm.
Study next
Common traps
- Answering 24, the half-chord, without doubling it
- Subtracting the radii to get 16 rather than subtracting their squares
- Assuming the chord runs through the centre, which would make it the 52 cm diameter
The concentric pair works cleanly because only R² − r² is ever needed, so the two radii never have to be convenient numbers on their own.
24 Sep 2024, 16:00, Quant Q.16 runs it backwards: one line cuts both circles in chords of 6 cm and 18 cm and asks for R² − r². Circle-angle work sits later in this 12:30 paper at Quant Q.24.
Related PYQs
No directly related past PYQ was found.