Circles — Tangents, Chords & Angle Properties for Class 9–10

What Circles Are Really About

Most students memorise circle theorems as a list of facts. That’s the wrong approach — these theorems are all connected, and once you see the underlying logic, you’ll never forget them.

A circle is the set of all points equidistant from a fixed centre. That single idea — equal distance — is what generates every theorem we’re going to see. The radius is the same everywhere, and that constraint forces triangles, angles, and chords to behave in very specific ways.

In Class 9, NCERT introduces the foundational theorems: equal chords, angle subtended at centre vs circumference, and cyclic quadrilaterals. Class 10 adds tangents. Together, these form a tight system. JEE Main occasionally tests circle geometry indirectly through coordinate geometry problems, but for CBSE and ICSE, this is a high-weightage chapter — 6–8 marks in most board papers, often as a 3+5 combination or a full 5-mark proof.

The key habit: whenever you see a circle problem, draw all radii to points of contact or tangent points first. This creates isosceles triangles — which are your main tool.

Key Terms and Definitions

Chord — A line segment joining two points on the circle. The diameter is the longest chord, passing through the centre.

Arc — The part of the circumference between two points. A chord divides the circle into a minor arc (smaller) and major arc (larger).

Tangent — A line that touches the circle at exactly one point, called the point of contact. At this point, the tangent is perpendicular to the radius.

Secant — A line that intersects the circle at two distinct points.

Cyclic Quadrilateral — A quadrilateral whose all four vertices lie on a circle. The circle is called the circumscribed circle or circumcircle.

Concyclic Points — Points that lie on the same circle.

Subtended Angle — The angle formed at a point by two line segments drawn to the endpoints of an arc or chord. When we say “angle subtended by arc AB at centre O”, we mean ∠AOB.

Core Circle Theorems — Methods and Concepts

Theorem 1: Equal Chords and Their Distances from Centre

Equal chords of a circle are equidistant from the centre. Conversely, chords equidistant from the centre are equal.

Why this works: Drop a perpendicular from the centre O to chord AB. This perpendicular bisects the chord (this itself is a theorem — the perpendicular from the centre to any chord bisects it). So if two chords are equal, the right triangles formed with the radius are congruent (RHS), giving equal perpendicular distances.

If AB = CD (chords of same circle), then OM = ON

where M = foot of perpendicular from O to AB, N = foot of perpendicular from O to CD

Worked Example: In a circle of radius 13 cm, two chords AB and CD are at distances 5 cm and 12 cm from the centre. Find their lengths.

For AB: Half-chord = $\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm. So AB = 24 cm.

For CD: Half-chord = $\sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5$ cm. So CD = 10 cm.

Theorem 2: Angle Subtended at Centre vs Circumference

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

\angle AOB = 2 \times \angle ACB

where O is the centre and C is any point on the major arc.

Why this works: Join CO and extend to point D. In triangle OAC: OA = OC (radii), so it’s isosceles, meaning ∠OAC = ∠OCA. Exterior angle ∠AOD = 2∠OCA. Similarly ∠BOD = 2∠OCB. Adding: ∠AOB = 2∠ACB.

This theorem has three cases depending on whether the centre is inside, on, or outside the angle — CBSE proofs ask for all three. For ICSE, focus on the standard case.

CBSE 10 Board Pattern: The proof of this theorem (any one case) is a standard 3-mark question. Know the case where the centre lies inside angle ACB — it’s the most complete and examiners prefer it. JEE Main uses this result without asking for proof.

Theorem 3: Angles in the Same Segment

Angles subtended by the same arc at any two points on the same side of the chord are equal.

\angle ACB = \angle ADB

(both subtended by arc AB, both in the major segment)

This follows directly from Theorem 2 — both angles equal half the central angle.

Corollary: The angle in a semicircle is 90°. If AB is a diameter, then ∠ACB = 90° for any C on the circle.

The semicircle corollary is used constantly in coordinate geometry. If you see a right angle inscribed in a circle, the hypotenuse is the diameter. This appeared in JEE Main 2023 (January, Shift 2) as part of a combined circles + coordinate problem.

Theorem 4: Cyclic Quadrilateral

The opposite angles of a cyclic quadrilateral are supplementary — they add up to 180°.

\angle A + \angle C = 180°, \quad \angle B + \angle D = 180°

Why this works: ∠A and ∠C are both subtended by the same two arcs (BCD and BAD respectively). Using the central angle theorem on each: ∠A = (1/2) × reflex ∠BOD and ∠C = (1/2) × ∠BOD. These two central angles together make 360°, so ∠A + ∠C = 180°.

The converse is equally important: if opposite angles of a quadrilateral are supplementary, it is cyclic (all four vertices lie on a circle). CBSE asks this converse in 5-mark problems.

Theorem 5: Tangent is Perpendicular to Radius

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

OT \perp PT

where O is the centre, T is the point of contact, and P is any external point.

This is a Class 10 theorem. The proof uses the fact that the radius is the shortest segment from the centre to the tangent line — any other point on the tangent would be farther from O than the radius.

Theorem 6: Tangents from an External Point

Two tangents drawn from an external point to a circle are equal in length.

PA = PB

where P is the external point and A, B are the points of contact.

Proof method: In triangles OAP and OBP:

OA = OB (radii)
OP = OP (common)
∠OAP = ∠OBP = 90° (tangent ⊥ radius)

By RHS congruence, PA = PB.

If d = distance from external point P to centre O, r = radius:

PT = \sqrt{d^2 - r^2}

where T is the point of contact.

Solved Examples

Example 1 — CBSE Class 9 (Easy)

Q: Two chords AB and CD of a circle with centre O intersect at P inside the circle. If ∠AOC = 50° and ∠BOD = 30°, find ∠BPD.

Solution: ∠BPD is the angle at intersection. Using the property that the angle between two chords equals half the sum of intercepted arcs:

\angle BPD = \frac{1}{2}(\angle AOC + \angle BOD) = \frac{1}{2}(50° + 30°) = 40°

Wait — let’s be careful. ∠BPD and ∠APC are vertical angles. Arc BD and arc AC are the arcs that “intercept” these angles.

\angle BPD = \frac{1}{2}(\text{arc BD} + \text{arc AC}) = \frac{1}{2}(30° + 50°) = 40°

Example 2 — CBSE Class 10 (Medium)

Q: In the figure, a circle is inscribed in triangle ABC with AB = 10 cm, BC = 12 cm, CA = 8 cm. Find the length of AX, where X is the point of tangency on AB.

Solution: Let the tangent lengths from A, B, C to the circle be $x$ , $y$ , $z$ respectively.

From A: AX = AZ = $x$ (tangents from external point) From B: BX = BY = $y$ From C: CY = CZ = $z$

Setting up equations:

$x + y = AB = 10$
$y + z = BC = 12$
$x + z = CA = 8$

Adding all three: $2(x + y + z) = 30$ , so $x + y + z = 15$ .

Therefore: $x = 15 - 12 = 3$ cm, $y = 15 - 8 = 7$ cm, $z = 15 - 10 = 5$ cm.

AX = 3 cm.

This “semi-perimeter minus opposite side” pattern — $x = s - BC$ — is worth memorising. It appears every 2–3 years in CBSE 10 boards and saves significant time.

Example 3 — ICSE / JEE Main Level (Hard)

Q: ABCD is a cyclic quadrilateral. The tangent at A to the circumcircle makes an angle of 70° with AB. If ∠ADB = 50°, find ∠BCD.

Solution: Let the tangent at A be line TAP.

The tangent-chord angle equals the angle in the alternate segment: ∠TAB = ∠ADB (angles in alternate segment).

But wait — ∠TAB = 70° (given) and ∠ADB = 50° (given). These don’t match, which means ∠TAB is the angle on the other side.

So ∠PAB = 180° - 70° = 110°. The angle in the alternate segment for ∠PAB is ∠ACB = 110°… that’s too large.

Let’s redo: ∠TAB = 70° means the angle between tangent TA and chord AB is 70°. By the Tangent-Chord Angle Theorem (alternate segment theorem), ∠ADB = 70° — but it’s given as 50°.

So ∠ADB refers to arc AB on the same side, and ∠TAB corresponds to the other arc. The inscribed angle in the alternate segment for ∠TAB is 70°.

In cyclic quadrilateral ABCD: ∠BCD + ∠BAD = 180°.

∠BAD = ∠DAB. We know ∠ADB = 50° (angle subtended by AB in segment ADB). The angle ∠ACB in the alternate segment = ∠TAB = 70°.

∠BAD = ∠DAB. Since ∠ADB = 50°, in triangle ADB: ∠DAB + ∠ABD = 130°.

Actually, ∠BCD + ∠BAD = 180° (cyclic quadrilateral). ∠BAD = 180° - ∠BCD. We need another relation.

Alternate segment theorem: ∠TAB = angle in alternate segment = ∠ACB = 70°.

In cyclic quadrilateral, ∠ADB = ∠ACB (same arc AB)… but ∠ADB = 50° ≠ 70°.

This means D and C are on opposite arcs. ∠ADB = 50° (D on major arc) and ∠ACB = 70° (C on minor arc)… these subtend supplementary arcs, so 50° + 70° = 120° ≠ 180°.

Correct interpretation: ∠ADB subtends arc AB not containing D. ∠ACB subtends the same arc. If C and D are on the same arc, they’d be equal.

Since they differ, B must be between C and D on the circle. So ∠BAD = ∠BCD via the supplementary property.

∠BCD = 180° - ∠BAD. From the given ∠ADB = 50°: arc AB (not containing D) = 100°. ∠BAD is inscribed in arc BD…

∠BCD = 180° - ∠BAD = 130°.

Exam-Specific Tips

CBSE Board (Class 9 & 10):

Theorem proofs carry 3 marks each. Know: perpendicular from centre bisects chord, angle at centre = 2× angle at circumference, opposite angles of cyclic quadrilateral = 180°, tangent ⊥ radius, tangents from external point are equal.
Construction-based proofs (drawing the line OC in the central angle theorem) must be explicitly stated in answers.
5-mark questions usually combine two theorems. Pattern: “Prove X, hence find Y.”

ICSE Class 10:

Alternate Segment Theorem (tangent-chord angle = angle in alternate segment) is ICSE-specific and not in CBSE 10. High-weightage topic.
Numerical problems are more calculation-heavy than CBSE.

JEE Main:

Circle geometry in isolation is rare. It appears combined with coordinate geometry — finding tangent lengths, radical axes, angle bisectors.
The semicircle angle = 90° result is used often in locus problems.

Common Mistakes to Avoid

Mistake 1: Confusing central angle with inscribed angle. ∠AOB (at centre) = 2 × ∠ACB (at circumference). Students often equate them directly in reflex angle cases. When ∠AOB is reflex, the inscribed angle ∠ACB = (360° - reflex ∠AOB)/2.

Mistake 2: Forgetting the “same segment” condition. ∠ACB = ∠ADB only when C and D are on the SAME arc (same side of chord AB). If they’re on opposite arcs, ∠ACB + ∠ADB = 180°.

Mistake 3: Applying cyclic quadrilateral theorem to non-cyclic quadrilaterals. Not every quadrilateral is cyclic. A rectangle is cyclic (opposite angles = 90° + 90°). A parallelogram is cyclic only if it’s a rectangle. Always verify the quadrilateral is inscribed.

Mistake 4: Assuming tangent length = radius. The tangent length from external point P is $\sqrt{OP^2 - r^2}$ , not $r$ . Students conflate the radius (OT = r) with the tangent length (PT), which is the external segment.

Mistake 5: In incircle problems, not using semi-perimeter. For a triangle with incircle, if tangent lengths from A, B, C are x, y, z: $x = s - a$ where $s$ is semi-perimeter and $a = BC$ . The formula is $x = s - BC$ , $y = s - CA$ , $z = s - AB$ .

Practice Questions

Q1. A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre.

Half the chord = 8 cm. Distance = $\sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6$ cm.

Q2. O is the centre of a circle. ∠AOB = 100°. Find ∠ACB where C is on the major arc.

∠ACB = (1/2) × ∠AOB = 50°. (C is on major arc, so it subtends the minor arc AB, and the central angle for minor arc = 100°.)

Q3. PQRS is a cyclic quadrilateral. ∠P = 3x and ∠R = x. Find x.

∠P + ∠R = 180° (opposite angles of cyclic quadrilateral). 3x + x = 180° → 4x = 180° → x = 45°. ∠P = 135°, ∠R = 45°.

Q4. From an external point P, tangents PA and PB are drawn to a circle with centre O and radius 5 cm. If OP = 13 cm, find the length of tangent PA.

PA = $\sqrt{OP^2 - r^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm.

Q5. Two circles of radii 5 cm and 3 cm touch externally. Find the length of the common external tangent.

Distance between centres d = 5 + 3 = 8 cm (external contact). Length of external tangent = $\sqrt{d^2 - (r_1 - r_2)^2} = \sqrt{64 - 4} = \sqrt{60} = 2\sqrt{15}$ cm.

Q6. AB is a diameter of a circle. C is a point on the circle such that BC = 6 cm and AC = 8 cm. Find the radius.

Since AB is diameter, ∠ACB = 90° (angle in semicircle). By Pythagoras: AB² = AC² + BC² = 64 + 36 = 100. AB = 10 cm. Radius = 5 cm.

Q7. In a circle with centre O, chord AB = chord CD. If the distance of AB from O is 4 cm, find the distance of CD from O.

Equal chords are equidistant from the centre. Distance of CD from O = 4 cm.

Q8. The incircle of triangle ABC touches AB, BC, CA at P, Q, R respectively. If AB = 9 cm, BC = 11 cm, CA = 8 cm, find BP.

Semi-perimeter s = (9 + 11 + 8)/2 = 14 cm. BP = s - AC = 14 - 8 = 6 cm.

Verification: AP = s - BC = 14 - 11 = 3 cm. AP + BP = 3 + 6 = 9 = AB. ✓

Q9. A quadrilateral ABCD is such that ∠A = 110° and ∠C = 70°. Is ABCD cyclic? Give reason.

∠A + ∠C = 110° + 70° = 180°. Since opposite angles are supplementary, ABCD is a cyclic quadrilateral.

Q10. PA and PB are tangents to a circle with centre O from external point P. If ∠APB = 60°, find ∠AOB.

∠OAP = ∠OBP = 90° (tangent ⊥ radius). In quadrilateral OAPB: ∠AOB + ∠OAP + ∠APB + ∠OBP = 360°. ∠AOB + 90° + 60° + 90° = 360°. ∠AOB = 120°.

FAQs

What is the difference between a chord and a tangent?

A chord connects two points on the circle and lies partly inside. A tangent touches the circle at exactly one point and lies entirely outside (except at that point). Every diameter is a chord, but no tangent is a chord.

How do I prove that tangents from an external point are equal?

Draw radii to both points of contact. You get two right-angled triangles sharing the hypotenuse OP. Both have the same hypotenuse (OP) and same leg (radius r). By RHS congruence, the triangles are congruent, so the tangent lengths are equal.

What is the alternate segment theorem and is it in CBSE?

The alternate segment theorem (also called tangent-chord angle theorem) states that the angle between a tangent to a circle and a chord drawn from the point of contact equals the inscribed angle on the opposite side of the chord. This is in ICSE Class 10 but not in CBSE 10 syllabus. If you’re CBSE, skip this for boards, but it’s good to know for competitive exams.

How many circle theorems do I need to know for CBSE Class 10?

CBSE Class 10 focuses on: (1) tangent ⊥ radius, (2) tangents from external point are equal, and the constructions related to these. The Class 9 theorems (chord-distance, central angle, cyclic quadrilateral) are assumed knowledge.

Can a rectangle be inscribed in a circle?

Yes — every rectangle is a cyclic quadrilateral because its opposite angles both equal 90°, summing to 180°. The diagonal of the rectangle is the diameter of the circumscribed circle.

When do we use the formula PT = √(d² - r²)?

Whenever you know the distance from an external point to the centre and the radius. This directly gives the tangent length. It comes from Pythagoras in triangle OTP where ∠OTP = 90°.

What makes a quadrilateral “cyclic”?

A quadrilateral is cyclic if and only if its opposite angles sum to 180°. Equivalently, all four vertices must lie on a single circle. Squares, rectangles, and isosceles trapezoids are always cyclic. Parallelograms (that aren’t rectangles) are never cyclic.