Equal Chords Are Equidistant from the Centre
A chord is a straight line segment whose endpoints lie on the circle. When two chords in the same circle have equal length, they are positioned symmetrically with respect to the centre.
This means the perpendicular distances from the centre of the circle to each chord are equal.
To understand why, consider two chords AB and CD of equal length in a circle with centre O. Draw perpendiculars OM and ON from O to chords AB and CD respectively, where M and N are the foot of the perpendiculars on the chords. Because AB = CD, triangles OMA and ONC are congruent by RHS (right angle, hypotenuse, side) criteria, implying OM = ON. Hence, equal chords lie at the same distance from the centre.
- Equal chords have equal lengths.
- Their perpendicular distances from the centre are equal.
- This property helps identify equal chords or distances in problems.
Perpendicular Bisector of a Chord Passes Through the Centre
The perpendicular bisector of a chord is a line that cuts the chord into two equal parts at a right angle.
In any circle, this line always passes through the centre of the circle.
For example, given chord AB, construct its perpendicular bisector. This line will intersect the circle's centre O. This occurs because the centre is equidistant from the endpoints A and B of the chord, so it lies on the locus of points equidistant from A and B — the perpendicular bisector of AB.
- Perpendicular bisector divides the chord into two equal segments.
- It is always a line through the circle's centre.
- Useful for locating the centre or proving points lie on the same circle.
Tangents from an External Point Are Equal in Length
A tangent to a circle is a straight line that touches the circle at exactly one point.
If two tangents are drawn from the same point outside the circle, these tangents have equal lengths from that point to the points of contact on the circle.
For example, from point P outside circle O, tangents PA and PB touch the circle at A and B respectively. Triangles OAP and OBP are congruent by RHS criteria (OA = OB as radii, OP common, and right angles at A and B because radius is perpendicular to tangent). Therefore, PA = PB.
- Tangents touch the circle at exactly one point.
- Tangents from the same external point are equal in length.
- This property is often used to prove equal lengths or angles in problems.
Applying Symmetry Properties to Solve Problems
These symmetry properties allow us to justify equal lengths, distances, and angles in circle geometry problems.
For instance, if two chords are equal, you can state their distances from the centre are equal without measurement.
If a line bisects a chord at right angles, you can conclude it passes through the centre, which helps in locating the centre or proving points are collinear with the centre of the circle.
- Use equal chord property to find unknown distances.
- Use perpendicular bisector property to locate the centre or prove collinearity.
- Use equal tangents property to prove lengths or angles are equal.
- Justify answers by referencing these properties explicitly.
Worked Example: Finding Distance from Centre to a Chord
In a circle with radius 10 cm, two chords AB and CD are equal in length, each 12 cm.
Find the distance from the centre O to each chord.
Solution: Let OM and ON be perpendicular distances from O to chords AB and CD respectively, with M and N on the chords. Since AB = CD = 12 cm, and radius OA = OC = 10 cm, consider triangle OMA where AM = 6 cm (half the chord length). Using Pythagoras: OM = √(OA² - AM²) = √(10² - 6²) = √(100 - 36) = √64 = 8 cm. Similarly, ON = 8 cm. Thus, the chords are 8 cm from the centre, confirming equal chords are equidistant from the centre.
- Calculate half chord length.
- Apply Pythagoras theorem in right triangle formed by radius, half chord, and distance from centre.
- Conclude equal chords have equal distances from centre.
At-a-glance summary
Comparison table
| Step | Detail | Example | Exam Use |
|---|---|---|---|
| Identify equal chords | Check if chords have the same length | Chords AB and CD both 12 cm | State equal chords are equidistant from centre to justify distances |
| Construct perpendicular bisector | Draw line at right angles bisecting chord | Bisector of chord AB passes through centre O | Use to prove centre lies on bisector or find centre |
| Draw tangents from external point | From point P outside circle, draw two tangents | Tangents PA and PB from P to circle | State tangents from same point are equal length to prove equal segments |
| Apply Pythagoras theorem | Use right triangle formed by radius, half chord, and distance from centre | Calculate distance OM = √(10² - 6²) = 8 cm | Show calculations clearly to find distances or lengths |
Active recall
Quick check
1. If two chords in a circle are equal in length, what can be said about their distances from the centre?
They are equal; equal chords are equidistant from the centre.
2. What is true about the perpendicular bisector of a chord in a circle?
It passes through the centre of the circle.
3. From a point outside a circle, two tangents are drawn. How do their lengths compare?
They are equal in length.
4. Why does the perpendicular bisector of a chord pass through the centre?
Because the centre is equidistant from the chord's endpoints, so it lies on the locus of points equidistant from those endpoints, which is the perpendicular bisector.