SSC CGL 2025 Tier-I · 23 September 2025, Shift 3 · Quantitative Aptitude · Question 22 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 13 September 2026.

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A line from the center of a circle bisects a chord. What is the angle between this line and the chord?

(a)30°
(b)45°
(c)90°
(d)60°

Answer

Why

Correct — C. Let O be the centre, AB the chord and M its midpoint, so AM = MB.

OA = OB (both are radii)
AM = MB (M bisects AB)
OM is common to △OAM and △OBM
So △OAM ≅ △OBM (SSS), and ∠OMA = ∠OMB

The two angles sit on the straight line AB, so each is 180° ÷ 2 = 90° → option (c)

Why the others are wrong

(a)
30° — 30° would make the angles at M 30° and 150°. The congruent triangles OAM and OBM force those two angles to be equal, so neither can be 30°.
(b)
45° — 45° would split the straight angle at M into 45° and 135°. SSS congruence of △OAM and △OBM makes them equal, so each is 90°.
(d)
60° — 60° leaves 120° on the other side of M. Equal radii and equal halves of the chord make the two angles equal, and two equal angles on a line are 90° each.

Concept

The chord–centre theorem runs both ways. The perpendicular from the centre to a chord bisects it, and the line from the centre to a chord's midpoint is perpendicular to it.

So the centre lies on the perpendicular bisector of every chord. That right angle at the midpoint is what turns chord problems into Pythagoras: (half-chord)² + (distance)² = r².

The theorem assumes the chord is not a diameter. The centre is itself the midpoint of a diameter, so any line through the centre passes through that midpoint at whatever angle it makes.

Key facts

The perpendicular from the centre of a circle to a chord bisects the chord.
The line joining the centre to the midpoint of a chord that is not a diameter is perpendicular to the chord.
Half-chord, distance from the centre and radius form a right triangle: (half-chord)² + d² = r².

Study next

Chord length from radius and distance: 2√(r² − d²)Equal chords are equidistant from the centreTwo parallel chords on the same side or on opposite sides of the centre

Common traps

Using the full chord instead of half the chord in (half-chord)² + d² = r² once this right angle is set up.
Confusing this angle with the angle the chord subtends at the centre, which changes with the chord's length.

The right angle at the midpoint drives chord-length items: 19 Sep 2024, 12:30, Quant Q.16 (radius 10 cm, distance 6 cm, chord = 2√(100 − 36) = 16 cm).

24 Sep 2025, 09:00, Quant Q.24 uses radius 13 cm and distance 5 cm: chord = 2√(169 − 25) = 24 cm.

Related PYQs

No directly related past PYQ was found.

Topic: Circle and its chords, tangents, angles subtended by chords of a circle, common tangents to two or more circles — all 140 SSC CGL questions on it, every solved shift.