Circles is one of the most theorem-rich and rewarding chapters in the CDS and OTA geometry section. A circle hides several elegant rules — equal chords, the tangent-radius right angle, angles in the same segment and cyclic quadrilaterals. Learn this small set of facts and you can crack two or three questions every year almost on sight, with no lengthy calculation.
Why Circles Is High-Value in CDS
In the CDS Elementary Mathematics paper you can expect 2 to 3 questions from Circles almost every year, and the General Science / geometry mix of OTA leans on the same ideas. The reason this topic rewards preparation is that it is built almost entirely on a fixed list of theorems. Once you know the theorem, the answer often follows in a single step — there is rarely heavy algebra.
Unlike triangles, where a problem may demand congruence proofs or long constructions, circle questions in CDS are usually direct applications: a tangent meets a radius, so the angle is 90°; an angle stands on a diameter, so it is 90°; opposite angles of a cyclic quadrilateral add to 180°. The examiner tests whether you remember the rule, not whether you can derive it.
This makes Circles a confidence chapter. The mensuration part — circumference, area and sectors — uses only two or three formulas, and the geometry part uses a compact theorem set. Master both halves and you bank reliable marks while spending very little time per question.
Always sketch the circle and mark the centre. The moment you draw the radius to a point of tangency or to a chord's midpoint, the right angle appears and most questions solve themselves.
The Vocabulary of a Circle
A circle is the set of all points in a plane that are at a fixed distance, the radius, from a fixed point called the centre. Get the terms exact, because CDS options often differ only in wording.
- Radius (r) — distance from the centre to any point on the circle.
- Diameter (d) — a chord passing through the centre; d = 2r, the longest chord.
- Chord — a line segment joining any two points on the circle.
- Secant — a line that cuts the circle at two points.
- Tangent — a line that touches the circle at exactly one point.
- Arc — a part of the circle between two points; the smaller is the minor arc, the larger the major arc.
- Segment — the region between a chord and its arc.
- Sector — the region between two radii and the arc they cut off, like a slice of pie.
A semicircle is exactly half a circle, bounded by a diameter and its arc. The angle in a semicircle — an angle whose arms end at the two ends of a diameter — is always a right angle. This single fact answers many CDS questions.
Chord Theorems You Must Know
Chords obey three clean rules that the examiner uses again and again.
1. The perpendicular drawn from the centre to a chord bisects the chord.
2. Conversely, the line joining the centre to the midpoint of a chord is perpendicular to the chord.
3. Equal chords are equidistant from the centre; and chords equidistant from the centre are equal.
The first rule is the workhorse. If a chord of length 2a is at distance p from the centre of a circle of radius r, drop the perpendicular from the centre to the chord. It splits the chord into two halves of length a, and a right triangle forms with legs a and p and hypotenuse r. So r2 = a2 + p2, which is just Pythagoras applied to the circle.
This lets you find a chord's length from its distance, or its distance from its length, in one step. Longer chords lie closer to the centre; the diameter, the longest chord of all, passes right through it at distance zero.
Tangents and Their Right Angles
A tangent touches a circle at one point only, and everything about tangents flows from one fact.
The tangent at any point of a circle is perpendicular to the radius drawn to the point of contact. So the radius and tangent meet at 90°.
From an external point you can draw exactly two tangents to a circle, and they have equal length. If P is outside the circle with centre O, and the tangents touch at A and B, then PA = PB. The line PO bisects the angle between the two tangents and also bisects the chord AB at right angles.
Because the radius-tangent angle is 90°, the length of a tangent from an external point at distance L from the centre is √(L2 − r2). This Pythagoras result is a frequent CDS shortcut.
There is also the alternate segment theorem: the angle between a tangent and a chord drawn from the point of contact equals the inscribed angle in the alternate segment on the other side of the chord. It looks technical but simply lets you transfer an awkward tangent angle into an easier inscribed angle.
Angles Subtended by Arcs and Chords
These angle rules turn many CDS items into one-line answers.
1. The angle subtended by an arc at the centre is twice the angle it subtends at any point on the remaining circle: ∠centre = 2 × ∠circumference.
2. Angles in the same segment are equal — every inscribed angle standing on the same arc has the same measure.
3. The angle in a semicircle is 90° (a special case of rule 1, where the central angle is 180°).
Rule 1 is the central-angle theorem, the parent of the whole set. If an arc makes 80° at the centre, it makes 40° at the circumference. Rule 2 follows immediately: any two points on the major arc see the same chord under the same angle, so those inscribed angles are equal.
Rule 3 is the most used of all in CDS. Whenever you see a triangle inscribed in a circle with one side passing through the centre — that is, a diameter — the angle opposite the diameter is a right angle. Spotting this saves the whole calculation.
Cyclic Quadrilaterals
A cyclic quadrilateral is a four-sided figure whose four vertices all lie on a single circle. It carries two neat properties.
1. The opposite angles of a cyclic quadrilateral are supplementary: each pair adds up to 180°.
2. The exterior angle of a cyclic quadrilateral equals the interior opposite angle.
So if one angle of a cyclic quadrilateral is 70°, the angle opposite it must be 110°. This is the most common circle-angle question in CDS, and it is answered by a single subtraction from 180°.
If a question gives you a quadrilateral inscribed in a circle and three angles, you can find the fourth instantly using the supplementary-opposite-angles rule — no need to chase arcs or central angles.
The converse is also examined: if the opposite angles of a quadrilateral sum to 180°, the quadrilateral can be inscribed in a circle. A rectangle is always cyclic; a non-square rhombus never is, because its opposite angles are equal, not supplementary.
Circumference, Area and Sectors
The measurement half of the chapter rests on just a few formulas. Take π ≈ 22÷7 or 3.14, as the question prefers.
Circumference = 2πr = πd
Area of circle = πr2
Length of an arc of angle θ° = (θ ÷ 360) × 2πr
Area of a sector of angle θ° = (θ ÷ 360) × πr2
Area of a ring (annulus) = π(R2 − r2)
Notice that arc length and sector area are simply fractions of the whole circle, the fraction being θ÷360. A 90° sector is a quarter circle; a 60° sector is one-sixth. This proportional thinking is faster than memorising separate formulas.
The area of a sector can also be written as ½ × r × (arc length). And a semicircle's perimeter is πr + 2r, not just πr — you must add the diameter.
Worked Example: Chord Length from Distance
A chord is at a distance of 8 cm from the centre of a circle whose radius is 17 cm. Find the length of the chord.
So the chord is 30 cm long. The whole solution is one application of the perpendicular-from-centre theorem followed by Pythagoras — the typical depth of a CDS circle question.
Common Mistakes That Cost Marks
Forgetting that the perpendicular from the centre bisects the chord. Students use the full chord length as a leg in Pythagoras instead of half the chord, doubling their error.
Confusing the central angle with the inscribed angle. The angle at the centre is twice the angle at the circumference, not equal to it. Mixing these up halves or doubles the answer.
In a cyclic quadrilateral, adding adjacent angles to 180° instead of opposite ones. Only opposite angles are supplementary.
One more frequent slip: forgetting to add the diameter when finding the perimeter of a semicircle, or using πd when the question asks for area. Read whether circumference or area is wanted before you pick the formula.
Previous-Year Style Question
Q. Two tangents are drawn from an external point P to a circle with centre O, touching it at A and B. If ∠APB = 50°, what is the measure of ∠AOB?
Answer: OA ⊥ PA and OB ⊥ PB because a radius is perpendicular to the tangent at the point of contact, so ∠OAP = ∠OBP = 90°. In quadrilateral OAPB, the four angles sum to 360°: ∠AOB + ∠OAP + ∠APB + ∠OBP = 360°. Thus ∠AOB + 90° + 50° + 90° = 360°, giving ∠AOB = 360° − 230° = 130°. (Quick check: ∠AOB and ∠APB are supplementary, 130° + 50° = 180°.)
A Quick Method Checklist
Run this fixed sequence on every circle question to stay fast and accurate:
- Draw the circle and mark the centre O clearly.
- For chord problems, drop the perpendicular from the centre and use r2 = a2 + p2.
- For tangents, mark the 90° radius-tangent angle and use equal tangent lengths.
- For angles, decide if you need the central-angle (double), same-segment (equal), or semicircle (90°) rule.
- For quadrilaterals on the circle, use opposite angles = 180°.
- For measurement, pick circumference, area, arc or sector and confirm π as 22÷7 or 3.14.
If a figure shows a triangle inscribed with one side as the diameter, write 90° at the opposite vertex before doing anything else — it is almost always the key to the answer.
Quick Revision
- Diameter = 2r is the longest chord; perpendicular from centre bisects any chord.
- Chord rule: r2 = (half chord)2 + (distance from centre)2.
- Tangent ⊥ radius at the point of contact; two tangents from a point are equal.
- Tangent length from an external point = √(L2 − r2).
- Central angle = 2 × inscribed angle; angle in a semicircle = 90°.
- Angles in the same segment are equal.
- Cyclic quadrilateral: opposite angles add to 180°; exterior angle = interior opposite.
- Circumference = 2πr, area = πr2, sector area = (θ÷360) × πr2.
Frequently asked questions
How many questions come from Circles in CDS?
Usually 2 to 3 questions appear across the geometry and mensuration parts of the Elementary Mathematics paper. Because they rest on a fixed set of theorems, they are highly scoring once those rules are memorised.
What is the single most useful circle theorem for CDS?
The angle in a semicircle is 90°. Whenever a triangle is inscribed with one side as the diameter, the opposite angle is a right angle, which cracks the problem instantly. Close behind is the tangent-perpendicular-to-radius rule.
How do I find a chord's length from its distance to the centre?
Drop a perpendicular from the centre to the chord; it bisects the chord. Then use r2 = (half chord)2 + (distance)2, a direct Pythagoras step, and double the half-chord to get the full length.
What is special about a cyclic quadrilateral?
Its four vertices lie on one circle, and its opposite angles always add up to 180°. Also, an exterior angle equals the interior opposite angle. These two facts answer most circle-angle questions in a single subtraction.
Which value of pi should I use in the exam?
Use 22÷7 when the radius or diameter is a multiple of 7, since the working stays clean, and 3.14 otherwise. Always check whether the question asks for circumference or area before choosing the formula.
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