Circles
~13 min read
- What: Chord, tangent, secant, arc, sector, segment. Angle subtended by an arc / chord at centre vs. circumference. Tangent-radius perpendicularity. Length of tangents from an external point. Cyclic quadrilateral.
- Why it matters: CDS averages 4–6 questions per paper on this chapter. Many properties; once memorised, the questions resolve in seconds.
- Key fact: The angle subtended by a chord at the centre is twice the angle subtended at any point on the major arc (the inscribed angle theorem). Inscribed angles standing on the same arc are equal.
Circles is the most "geometry-flavoured" chapter in CDS Maths. The properties are visual — drawing the diagram correctly is half the work. The other half is recognising which of the dozen circle theorems applies. Master both and you collect 4–6 marks per paper.
This page is built from CDS Previous Year Questions across 2000–2023, plus NCERT Class 9 Curves and Circles and Class 10 Circles. Pair with Area and Perimeter, Lines and Angles, and Triangles.
What This Topic Covers
CDS scope: (1) basic terms — radius, diameter, chord, arc, secant, tangent; (2) inscribed angle theorem — central angle = 2 × inscribed angle; (3) angles in same segment; (4) cyclic quadrilateral — opposite angles supplementary; (5) tangent properties — perpendicular to radius, equal tangents from external point; (6) chord properties — perpendicular from centre bisects, equal chords equidistant; (7) secant-tangent / chord-chord power of point.
Why This Topic Matters
- 4–6 CDS questions per paper.
- Strong cross-link with Area and Perimeter (sector area) and Triangles (chord triangles).
- Tangent-radius perpendicularity is a 5-second test.
Exam Pattern & Weightage
| Year / Paper | No. | Subtopics Tested |
|---|---|---|
| 2007-II | 3 | Chord, central angle |
| 2008-I/II | 4 | Tangent, inscribed angle |
| 2009-II | 4 | Cyclic, tangent properties |
| 2010-I/II | 4 | Mixed |
| 2011-I/II | 5 | Sector, tangent length |
| 2013-I/II | 5 | Inscribed angle, chord |
| 2014-I/II | 4 | Mixed |
| 2015-I/II | 5 | Cyclic quadrilateral |
| 2016-I/II | 4 | Mixed |
| 2017-I/II | 5 | Tangent-radius, chord |
| 2018-I/II | 4 | Mixed |
| 2019-II | 4 | Mixed |
| 2020-I/II | 5 | Mixed |
| 2021-I/II | 5 | Inscribed, tangent |
| 2022-I / 2023-I | 5 | Mixed |
The angle subtended by the diameter at any point on the circle is always 90° (special case of the inscribed angle theorem with central angle 180°). This single fact solves dozens of "find the angle" CDS problems.
Core Concepts
Basic Terms
Chord: line segment with both endpoints on the circle. Diameter: longest chord, passes through centre. Tangent: line touching the circle at exactly one point. Secant: line intersecting the circle at two points. Arc: portion of the circle between two points. Sector: region between two radii and the arc. Segment: region between a chord and the arc.
Inscribed Angle Theorem
Two angles in the same segment are equal.
Cyclic Quadrilateral
Conversely: if opposite angles are supplementary, the quadrilateral is cyclic.
Tangent Properties
Two tangents from an external point are equal in length.
\(PT^2 = OP^2 - r^2\) (Pythagoras on right triangle of radius, tangent, line from external point).
Chord Properties
Equal chords are equidistant from the centre.
Length of chord at distance \(d\) from centre: \(2\sqrt{r^2 - d^2}\).
Power of a Point
A tangent is perpendicular to the radius at the point of contact, not at the centre. Always draw the radius to the tangent point, then mark the right angle there.
Worked Examples
Example 1 — Angle in Semicircle (2010-I)
Q: In a circle, \(AB\) is the diameter. \(C\) is a point on the circle. Find \(\angle ACB\).
- Diameter subtends 90° at any point on the circle (angle in a semicircle).
- \(\angle ACB = 90°\).
Example 2 — Central vs Inscribed (2013-I)
Q: A chord subtends 70° at the centre. Find the angle subtended at a point on the major arc.
- Inscribed angle = (central angle)/2 = 35°.
Example 3 — Tangent Length (2011-II)
Q: A circle has radius 5. From an external point at distance 13 from the centre, find the tangent length.
- Pythagoras: \(PT^2 = 13^2 - 5^2 = 169 - 25 = 144 \implies PT = 12\). (5-12-13 triple.)
Example 4 — Cyclic Quadrilateral (2015-I)
Q: In a cyclic quadrilateral \(ABCD\), \(\angle A = 70°\) and \(\angle B = 110°\). Find \(\angle C\) and \(\angle D\).
- Opposite supplementary: \(\angle A + \angle C = 180° \implies \angle C = 110°\). And \(\angle B + \angle D = 180° \implies \angle D = 70°\).
Example 5 — Chord Length (2009-II)
Q: A circle has radius 10. A chord is at perpendicular distance 6 from the centre. Find the chord's length.
- Length \(= 2\sqrt{r^2 - d^2} = 2\sqrt{100 - 36} = 2 \cdot 8 = 16\).
Example 6 — Power of Point Chords (2017-II)
Q: Two chords \(AB\) and \(CD\) intersect at \(P\). \(PA = 3, PB = 8, PC = 4\). Find \(PD\).
- \(PA \cdot PB = PC \cdot PD \implies 3 \cdot 8 = 4 \cdot PD \implies PD = 6\).
Example 7 — Tangents from External Point (2014-II)
Q: Two tangents from an external point to a circle of radius 7 cm are 24 cm each. Find the distance from the external point to the centre.
- By Pythagoras in the right triangle (radius, tangent, line to external point): distance \(= \sqrt{7^2 + 24^2} = \sqrt{625} = 25\) cm. (7-24-25 triple.)
How CDS Tests This Topic
Seven recurring archetypes: (1) angle in semicircle = 90°, (2) central vs inscribed angle (\(2\theta\) vs \(\theta\)), (3) tangent perpendicular to radius, (4) tangent length via Pythagoras, (5) cyclic quadrilateral opposite supplementary, (6) chord length from distance to centre, (7) power of point (chord-chord or secant-tangent).
Exam Shortcuts (Pro-Tips)
Shortcut 1 — Angle in Semicircle = 90°
If you see "diameter" in a CDS problem and an angle on the circle, that angle is 90°. Skip the calculation.
Shortcut 2 — Central = 2 × Inscribed
Same arc: central angle is exactly twice the inscribed angle. Apply directly.
Shortcut 3 — Pythagorean Triples Again
Tangent/radius/external-point right triangles often use 3-4-5, 5-12-13, 7-24-25, 8-15-17, 20-21-29. Recognise instantly.
Shortcut 4 — Tangents from External Point
Equal in length. They also subtend equal angles at the centre.
Shortcut 5 — Chord Length Formula
\(\text{chord} = 2\sqrt{r^2 - d^2}\) where \(d\) is the perpendicular distance from centre. Used in many CDS chord problems.
Common Question Patterns
Pattern 1 — Inscribed Angle Theorem
Given central angle, find inscribed (or vice versa). \(\text{inscribed} = \text{central}/2\).
Pattern 2 — Tangent Length
Right triangle of radius and tangent. Pythagoras.
Pattern 3 — Cyclic Quadrilateral
Opposite supplementary. Find missing angle.
Pattern 4 — Chord Length / Distance
Apply \(\text{chord} = 2\sqrt{r^2 - d^2}\) or rearrange.
Pattern 5 — Power of Point
Two chords intersecting inside, or secant-tangent from outside. Apply the product identity.
Preparation Strategy
Week 1. Master all chord and tangent properties. Drill 20 problems on inscribed angle theorem and tangent-radius perpendicularity. Memorise Pythagorean triples (they show up everywhere).
Week 2. Cyclic quadrilateral, power of point, sector and segment areas. Cross-train with Area and Perimeter for sector formulas.
Mock testing. Use CDS mock tests. Most slip-ups: missing the diameter ⇒ 90° trick, or confusing inscribed and central angles.
Drill Circles at Speed
CDS mocks with inscribed angle, tangent length, cyclic quadrilateral, and power-of-point problems. Seven archetypes — clean theorems — reflex.
Start Free Mock TestFrequently Asked Questions
What's the inscribed angle theorem?
The angle subtended by an arc at the centre is twice the angle subtended at any point on the major arc. Or: inscribed angle = (central angle)/2 if they stand on the same arc.
Why is the angle in a semicircle always 90°?
Because the diameter subtends a central angle of 180°. The corresponding inscribed angle is 180°/2 = 90°. Special case of the inscribed angle theorem.
What's the rule for tangents from an external point?
Two tangents from an external point are equal in length. They also subtend equal angles at the centre. The line from the external point to the centre bisects the angle between the tangents.
How do I find the length of a tangent from an external point?
Apply Pythagoras to the right triangle formed by the radius (to the point of contact), the tangent, and the line from the centre to the external point. \(PT^2 = OP^2 - r^2\). Pythagorean triples are very common.
What is "power of a point"?
For a point \(P\) and a circle: if a chord through \(P\) cuts the circle at \(A, B\), then \(PA \cdot PB\) is a constant (the power of \(P\)). For two chords through \(P\): \(PA \cdot PB = PC \cdot PD\). For a secant and tangent from outside: \(PA \cdot PB = PT^2\).
How do I find the length of a chord at known distance from centre?
For a chord at perpendicular distance \(d\) from the centre of a circle of radius \(r\): length = \(2\sqrt{r^2 - d^2}\). The perpendicular from centre bisects the chord.
Which CDS Maths topics connect to Circles?
Area and Perimeter — sector, segment, full area. Lines and Angles — angle chasing. Triangles — chord triangles. Cyclic quadrilateral.