Circles

Circles are everywhere in aptitude exams — from finding areas of shaded regions to tangent-based problems. The trick with circles is that there are a handful of theorems, and once we nail those down, every problem becomes a plug-and-solve situation. Let’s get into it.

Circle Basics

A quick refresher on the terminology:

• Radius (r): Distance from center to any point on the circle
• Diameter (d): Distance across the circle through the center (d = 2r)
• Chord: A line segment connecting any two points on the circle (diameter is the longest chord)
• Arc: A part of the circumference (minor arc < semicircle, major arc > semicircle)
• Sector: The “pizza slice” area between two radii and an arc
• Segment: The area between a chord and the arc it cuts off

Key Formulas

In simple language: for arc length and sector area, we just take that fraction of the full circle. If the angle is 60°, that’s 60/360 = 1/6 of the full circumference or area.

Circle Theorems — The Must-Know Set

1. Perpendicular from Center to Chord

A perpendicular drawn from the center of a circle to a chord bisects the chord (cuts it in half). This is incredibly useful — whenever we see a chord problem, we drop a perpendicular from the center and form a right triangle.

2. Equal Chords

Equal chords are equidistant from the center. The only difference is, the converse is also true — chords equidistant from the center are equal.

3. Tangent Properties

• A tangent is always perpendicular to the radius at the point of contact (this creates a right angle — use Pythagoras!)
• Two tangents drawn from an external point are equal in length
• The angle between two tangents from an external point and the angle subtended at the center are supplementary (add up to 180°)

4. Inscribed Angle Theorem (The Star Theorem)

The angle subtended by an arc at the center is twice the angle subtended at any point on the remaining circle.

In simple language: if we draw an angle from an arc to the center, and another angle from the same arc to a point on the circle, the center angle is exactly double. So the inscribed angle = ½ × central angle.

Special case: An angle in a semicircle is always 90° (since the central angle is 180°, and half of that is 90°). This is a classic exam question.

5. Cyclic Quadrilateral

A cyclic quadrilateral is a four-sided figure where all four vertices lie on a circle.

Key property: Opposite angles of a cyclic quadrilateral add up to 180°.

So if ∠A + ∠C = 180° and ∠B + ∠D = 180°, it’s cyclic.

6. Angles in the Same Segment

Angles subtended by the same arc (or chord) at points on the same side of the chord are equal. Think of it like this — no matter where we put the point on the same arc, the angle stays the same.

Shortcut Methods and Tricks

Trick 1: The right triangle in chord problems Whenever a chord problem gives us the chord length and distance from center (or radius), immediately draw the perpendicular from center to chord. We get a right triangle with:

• Hypotenuse = radius
• One leg = half the chord length
• Other leg = distance from center to chord

Trick 2: Two tangent problems When two tangents are drawn from an external point, the figure formed (two tangents + two radii) creates a kite shape. The tangent lengths are equal, and we can use this symmetry to solve quickly.

Trick 3: Angle in semicircle = 90° If a question says “diameter subtends an angle at a point on the circle,” the answer is always 90°. Don’t even calculate.

Trick 4: Quick sector calculations For common angles, memorize these fractions:

• 60° = 1/6 of circle
• 90° = 1/4 of circle
• 120° = 1/3 of circle
• 180° = 1/2 of circle

Worked Examples

Example 1: Arc Length and Sector Area

Find the arc length and sector area for a circle with radius 14 cm and central angle 90°.

Arc length = (90/360) × 2π(14) = (1/4) × 28π = 7π = 22 cm (using π = 22/7)

Sector area = (90/360) × π(14)² = (1/4) × 196π = 49π = 154 cm²

Example 2: Chord and Distance

A chord of length 24 cm is drawn in a circle of radius 13 cm. Find the distance of the chord from the center.

Drop a perpendicular from center to chord. It bisects the chord, so half-chord = 12 cm.

Now we have a right triangle: radius = 13, half-chord = 12, distance = ?

distance² = 13² - 12² = 169 - 144 = 25

Distance = 5 cm

Notice: 5-12-13 is a Pythagorean triplet! Spotting these saves calculation time.

Example 3: Tangent Length

From a point 17 cm away from the center of a circle of radius 8 cm, a tangent is drawn. Find the tangent length.

The tangent is perpendicular to the radius at the point of contact, so we have a right triangle:

• Hypotenuse = distance from external point to center = 17
• One leg = radius = 8
• Other leg = tangent length = ?

Tangent² = 17² - 8² = 289 - 64 = 225

Tangent length = 15 cm

(Another triplet: 8-15-17!)

Example 4: Inscribed Angle

A chord subtends an angle of 50° at a point on the major arc. Find the angle it subtends at the center.

By the inscribed angle theorem, the central angle is twice the inscribed angle.

Central angle = 2 × 50° = 100°

Example 5: Cyclic Quadrilateral

In a cyclic quadrilateral ABCD, ∠A = 70° and ∠B = 110°. Find ∠C and ∠D.

Opposite angles sum to 180°:

• ∠C = 180° - ∠A = 180° - 70° = 110°
• ∠D = 180° - ∠B = 180° - 110° = 70°

Common Exam Variations

1. Finding the area of a segment — sector area minus triangle area. The triangle is formed by two radii and the chord.
2. Two circles touching — externally (distance between centers = R + r) or internally (distance = R - r)
3. Common tangent problems — direct common tangents (don’t cross the line joining centers) vs transverse tangents (do cross it)
4. Shaded region between two concentric circles (ring/annulus area = π(R² - r²))
5. Semicircle problems — using the “angle in semicircle = 90°” property

Practice Problems

Q1: A sector has a central angle of 120° and radius 21 cm. Find its area and the length of the arc. (Use π = 22/7)

Q2: Two tangents PA and PB are drawn from a point P to a circle with center O and radius 5 cm. If OP = 13 cm, find the length of each tangent and the angle ∠APB.

Q3: In a circle of radius 10 cm, a chord is at a distance of 6 cm from the center. Find the length of the chord.

Answers

A1: Area = (120/360) × (22/7) × 21² = (1/3) × (22/7) × 441 = (1/3) × 1386 = 462 cm². Arc length = (120/360) × 2 × (22/7) × 21 = (1/3) × 132 = 44 cm.

A2: Tangent is perpendicular to radius, so PA² = OP² - OA² = 169 - 25 = 144. PA = 12 cm. In triangle OAP: sin(∠OPA) = 5/13, so ∠OPA = sin⁻¹(5/13) ≈ 22.6°. Since the figure is symmetric, ∠APB = 2 × 22.6° ≈ 45.2°. (Or we can note: ∠AOB + ∠APB = 180°, and tan(∠AOB/2) = 12/5, giving the same result.)

A3: Half-chord² = radius² - distance² = 100 - 36 = 64. Half-chord = 8 cm. Full chord = 16 cm.