The perpendicular from the center
Take a chord AB of a circle with center O. From O, draw the line that meets the chord at a right angle, and call the point where it meets the chord M. This line is the perpendicular from the center to the chord.
M is always the midpoint of the chord: AM = MB. The perpendicular from the center to a chord bisects the chord.
The perpendicular from the center O meets the chord AB at a right angle, at its midpoint.
Why: two congruent triangles
Join O to A and to B. The perpendicular OM splits the triangle OAB into two triangles, OMA and OMB. Compare them.
Both have a right angle at M. Their hypotenuses are OA and OB, and these are equal because both are radii. And the side OM belongs to both triangles. A right angle, the hypotenuse and one other side: the triangles are congruent by RHS.
The radii OA and OB, and the line from O at right angles to the chord, continued to the circle. It crosses AB at M and splits triangle OAB into two right-angled triangles, OMA and OMB.
So the halves are equal
In congruent triangles every part of one equals the matching part of the other. AM in triangle OMA matches MB in triangle OMB, so AM = MB: M is the midpoint of AB.
The same triangles give more. The angles AOM and BOM are equal, so the perpendicular also cuts the angle AOB at the center in half, and it passes through the middle of the arc AB.
The two ticked halves of the chord are equal.
The length of a chord, with Pythagoras
Triangle OMA has a right angle at M, so Pythagoras' theorem links the radius OA, the distance OM from the center to the chord, and the half-chord AM: OA² − OM².
A circle has a radius of 10 cm, and a chord is 6 cm from its center. Then , so AM = 8 cm. The chord is twice that: AB = 2 × 8 = 16 cm.
It works the other way too. A chord 24 cm long in a circle of radius 13 cm has halves of 12 cm, so its distance from the center is cm.
Finding the center of a circle
OA and OB are both radii, so the center is the same distance from A as from B, and the same is true for the two ends of every chord. The points equally far from A and from B are exactly the points of the perpendicular bisector of AB, the line through the midpoint of AB at right angles to it. So the perpendicular bisector of every chord passes through the center.
This finds the center of a circle when it is not marked. Draw any two chords that are not parallel, and construct the perpendicular bisector of each with a compass and a straightedge. Both lines pass through the center, and two lines cross at only one point, so the center is where they cross.
Two chords, AB and CD, each with its perpendicular bisector drawn across the circle. The two bisectors cross at the center.
Equal chords are equally far from the center
Two chords of the same length in one circle have equal halves. Their right-angled triangles have the same hypotenuse, the radius, and the same half-chord, so Pythagoras' theorem gives them the same third side: the two chords are the same distance from the center. In the circle of radius 10 cm, every chord 16 cm long is 6 cm from the center.
It works backwards as well: chords that are the same distance from the center are the same length. And the nearer a chord is to the center, the longer it is. The longest chord of all is 0 from the center: it is a diameter.
AB and CD are equal chords. The line from the center at right angles to each one meets it at its midpoint, and the two distances from the center to the chords are equal.
The usual mistakes
Stopping at half the chord. Pythagoras' theorem gives AM, from the midpoint to one end. The chord is 2 × AM.
Taking the depth for the distance to the center. In a round pipe, the depth of water is measured up from the bottom, which is a radius below the center. The distance from the center to the water surface is the radius minus the depth.
Dropping the perpendicular from a point that is not the center. Only the perpendicular from the center bisects the chord, because the two equal radii are what make the triangles congruent.
Worked example: The Width of the Water in a Round Drain Pipe
Question A drain pipe has a circular cross-section of radius 25 cm and center O. Water lies in the pipe to a depth of 10 cm at the deepest point. (a) How wide is the surface of the water? (b) After rain the water is 18 cm deep. How wide is the surface now?
1.Call the ends of the water surface A and B. The surface AB is a chord. Draw the perpendicular from O to AB, meeting it at M. The perpendicular from the center bisects the chord, so AM = MB.
The water surface AB is a chord. The perpendicular from the center O meets it at M and bisects it. 2.The lowest point of the pipe is 25 cm below O, and the water surface is 10 cm above the lowest point, so OM = 25 − 10 = 15 cm.
The bottom of the pipe is 25 cm below O and the water is 10 cm deep, so OM = 25 − 10 = 15 cm. 3.Triangle OMA is right-angled at M with hypotenuse OA = 25 cm. By Pythagoras' theorem, AM2 = 252 − 152 = 625 − 225 = 400, so AM = 20 cm.
Triangle OMA is right-angled at M: AM2 = 252 − 152 = 400, so AM = 20 cm. 4.(a) The surface is AB = 2 × 20 = 40 cm wide.
(a) The surface is AB = 2 × 20 = 40 cm wide. 5.When the depth is 18 cm, OM = 25 − 18 = 7 cm, so AM2 = 625 − 49 = 576 and AM = 24 cm. (b) The surface is now 2 × 24 = 48 cm wide. Check: 72 + 242 = 49 + 576 = 625 = 252.
(b) At 18 cm deep, OM = 7 cm, AM = √625 − 49 = 24 cm and the surface is 48 cm wide.
Answer: (a) 40 cm; (b) 48 cm
Common mistakes
- Using the depth 10 cm as the distance OM from the center to the surface. The depth is measured from the bottom of the pipe, and the center is a radius above the bottom, so OM = 25 − 10 = 15 cm.
- Giving 20 cm as the width. Pythagoras' theorem gives half the chord, from M to one end, so the full width is twice that.
More congruence, similarity and circle theorems problems, worked step by step →