Why do circles love tangents and chords?
Ever watched a bike wheel roll and notice the point where the tire just kisses the ground? That tiny kiss is a tangent – a line that touches a circle at exactly one point. And the line that cuts straight across a pizza slice? That’s a chord. Both hide neat tricks that pop up in every ICSE geometry problem.
💡 In Simple Words: A tangent just brushes a circle at one spot, while a chord is any straight line that joins two points on the circle. Knowing how they behave helps you solve many geometry questions quickly.
Key Definitions (first time, plain language)
- Tangent: A straight line that meets a circle at exactly one point. Think of it as a gentle tap on the circle’s surface.
- Chord: A straight line that connects two points on the circle’s edge. It’s like a bridge across the circle.
- Radius: The line from the centre of the circle to any point on the edge. It’s the circle’s “arm”.
- Diameter: A special chord that passes through the centre; it’s twice the radius.
Core Tangent Properties
1. Radius ⟂ Tangent at the point of contact
If you draw the radius to the point where the tangent touches, the two lines form a right angle (90°). Imagine a garden hose (radius) pushing straight out of a water pipe (tangent) – they’re perpendicular.
2. Tangents from an external point are equal
Pick a point outside the circle and draw two tangents to the circle. Both tangents have the same length. It’s like stretching two equal strings from a knot to touch a balloon – they’ll be the same length.
3. Angle between tangent and chord = angle in the alternate segment
The angle formed where a tangent meets a chord equals the angle formed by the chord with any other point on the circle, but on the opposite side of the chord. This is called the tangent‑chord theorem.
Core Chord Properties
1. Equal chords subtend equal angles at the centre
If two chords have the same length, the angles they make at the centre are equal. Think of two equally long slices of pizza – the central angles look the same.
2. Perpendicular from centre bisects a chord
A line drawn from the centre to the midpoint of a chord is perpendicular (forms a right angle) to that chord. It’s like a ruler dropping straight down to cut a piece of string exactly in half.
3. Distance from centre to chord formula
For a chord of length c in a circle of radius r, the distance d from the centre to the chord is given by d = √(r² – (c/2)²). This comes from the right‑triangle formed by the radius, half the chord, and the distance d.
Worked Example 1: Finding Tangent Length
Problem: From point P outside a circle with centre O and radius 5 cm, the distance OP is 13 cm. Find the length of the tangent PT where T is the point of contact.
Solution: Draw OT (radius) and PT (tangent). OT ⟂ PT, so OTP is a right‑angled triangle.
- Hypotenuse = OP = 13 cm
- One side = radius = 5 cm
- Use Pythagoras: PT² = OP² – OT² = 13² – 5² = 169 – 25 = 144
- Therefore, PT = √144 = 12 cm
Worked Example 2: Using the Tangent‑Chord Theorem
Problem: In a circle, a tangent at point A meets chord AB. If the angle between the tangent and chord AB is 40°, find the angle subtended by AB at any point on the opposite arc.
Solution: By the tangent‑chord theorem, the angle between the tangent and chord equals the angle in the alternate segment. Hence, the required angle is also 40°.
Worked Example 3: Distance from Centre to a Chord
Problem: A circle has radius 10 cm. A chord is 12 cm long. How far is the chord from the centre?
Solution:
- Half the chord = 6 cm
- Use formula d = √(r² – (c/2)²)
- d = √(10² – 6²) = √(100 – 36) = √64 = 8 cm
Quick Comparison Table
| Property | Applies to | Result |
|---|---|---|
| Radius ⟂ Tangent | Tangent at point of contact | Forms a right angle |
| Equal Tangents from external point | Two tangents from same outside point | Lengths are equal |
| Tangent‑Chord Theorem | Angle between tangent & chord | Equals angle in opposite arc |
| Equal chords ↔ equal central angles | Any two chords | Same length ⇔ same central angle |
| Perpendicular from centre bisects chord | Line from centre to chord midpoint | Right angle & chord split |
Tips for Solving ICSE Geometry Problems
- Always sketch the figure first – a clear diagram reveals right‑angles and equal lengths.
- Mark known radii, diameters, and right angles; they’re your shortcuts.
- When a problem mentions an “external point”, remember the two tangents from that point are equal.
- Use the chord‑distance formula whenever a chord length is given and you need the distance from the centre.
- For angle questions, decide whether the tangent‑chord theorem or the “angles in the same segment” rule applies.
📝 Likely Exam Questions
- Question: From a point 15 cm away from the centre of a circle of radius 9 cm, two tangents are drawn. Find the length of each tangent.
Answer: Using Pythagoras, tangent length = √(15² – 9²) = √(225 – 81) = √144 = 12 cm. - Question: In a circle, a chord of length 14 cm is 6 cm from the centre. Find the radius of the circle.
Answer: Using d² + (c/2)² = r² → 6² + 7² = r² → 36 + 49 = 85 → r = √85 cm. - Question: Prove that the angles subtended by two equal chords at the centre are equal.
Answer: Draw radii to the ends of each chord, forming two isosceles triangles with equal sides (radii) and equal bases (chords). By the Isosceles Triangle Theorem, the base angles – which are the central angles – are equal. - Question: The angle between a tangent and a chord at the point of contact is 30°. What is the angle in the alternate segment?
Answer: By the tangent‑chord theorem, the angle in the alternate segment is also 30°. - Question: Two chords AB and CD intersect at right angles at point E inside a circle of radius 10 cm. If AE = 6 cm and CE = 8 cm, find the length of the segment EB.
Answer: For intersecting chords, AE·EB = CE·ED. First find ED using Pythagoras in triangle OEC (O = centre). OE² = r² – CE² = 10² – 8² = 36 → OE = 6 cm. Since OE ⟂ CD, DE = √(r² – OE²) = √(100 – 36) = 8 cm. Now AE·EB = CE·ED → 6·EB = 8·8 → EB = 64/6 = 10.67 cm (approx).