Chord theorems
The center of a circle cannot be seen on the circle itself, but its chords can. What does a chord give away about where the center is, and how long the chord must be?
Where you are 15 in Circles
Why this exists
A circle is drawn around a center, but once it is drawn the center leaves no mark on it. A broken plate, the rim of a round table, the curve of a stone arch: each is a piece of a circle whose center is nowhere to be seen. What can be reached are chords, the straight segments between two points of the curve, because those can be measured with a ruler.
The theorems here tie each chord to the hidden center. The center always sits on the line that cuts a chord in half at right angles, so two chords are enough to find it. And how long a chord is depends on nothing but its distance from the center, so equal chords sit equally far out, equal chords cut equal arcs, and a chord, its arc and its distance from the center can each be worked out from the others.
Definition
Fix a circle with center O and radius r, and a Chords, secants and tangents \overline{AB}.
- The central angle of the chord is \angle AOB, the angle at the center between the radii to its ends. When \overline{AB} is a diameter this is a straight angle, 180^\circ.
- The arcs of the chord are the two arcs it cuts the circle into: the minor arc \overset{\frown}{AB}, measuring the same as \angle AOB, and the major arc, measuring 360^\circ minus that. When the chord is a diameter both are semicircles.
- The distance from the center to the chord is the length of the perpendicular segment from O to the line through A and B, the distance from a point to a line. Its foot is called M below; Result 1 shows that M is the midpoint of the chord. A diameter is at distance 0.
- Two circles are congruent when their radii are equal.
The theorems are about a line through the center, at right angles to a chord. Written with symbols: a line through O with \overleftrightarrow{OM} \perp \overline{AB} at M.
Intuition
Fold a paper circle along any line through its center and the two halves land exactly on each other, because every point of the circle is r from O and a fold through O keeps distances from O. So every line through the center is a line of mirror symmetry of the circle.
Now take a chord \overline{AB} and the line through O at right angles to it. Folding along that line sends A to a point of the circle on the same perpendicular to the fold, which can only be B. So the fold lays the chord over itself, half onto half, and lays the minor arc onto itself the same way. Everything that the fold swaps comes in equal pairs: AM = MB, the arc from A to C equals the arc from C to B, and the angle \angle AOM equals \angle BOM.
The same picture shows why distance decides length. Slide the chord toward the center, keeping it at right angles to the same diameter: its ends move out along the circle and the chord grows, reaching its longest, a diameter, when it passes through O. Two chords at the same distance from O are copies of each other turned about the center.
Results
Result 1 (the perpendicular from the center). If a line through the center is perpendicular to a chord, it bisects the chord. It also bisects the chord’s central angle, its minor arc and its major arc. For a chord at distance d from the center of a circle of radius r, the half-chord satisfies
Result 2 (the converses). (a) The line through the center and the midpoint of a chord that is not a diameter is perpendicular to the chord. (b) The perpendicular bisector of any chord passes through the center.
So the center of a circle is where the perpendicular bisectors of any two chords that are not parallel cross.
Result 3 (chords, central angles and arcs). In one circle, or in two congruent circles, the following say the same thing about two chords \overline{AB} and \overline{CD}, so each one forces the other two:
Here O is the center of each chord’s own circle and the arcs are the minor arcs.
Result 4 (chords and their distance from the center). In one circle, or in two congruent circles, two chords are equally long exactly when they are equally far from the center. Of two chords at different distances, the one nearer the center is the longer.
Why it is true
Result 1. Let the line through O be perpendicular to the chord \overline{AB} at M. If \overline{AB} is a diameter, the line meets it at O, which is its midpoint; the right angles it makes there halve the straight angle \angle AOB, and it meets the circle at the middle of each semicircle. Otherwise O is not on the chord, and the triangles OMA and OMB have
| Statements | Reasons |
|---|---|
| \angle OMA and \angle OMB are right angles | the line is perpendicular to \overline{AB} |
| OA = OB | both are radii |
| OM = OM | the same segment |
| \triangle OMA \cong \triangle OMB | hypotenuse leg congruence |
| AM = MB and \angle AOM \cong \angle BOM | matching parts of congruent triangles |
So M is the midpoint of the chord and the line bisects \angle AOB.
For the arcs, let the line meet the circle at C on the minor arc and at D on the major arc, so \overline{CD} is a diameter. The rays OC and OM are the same ray, so m\angle AOC = m\angle BOC, and a minor arc measures the same as its central angle, so m\overset{\frown}{AC} = m\overset{\frown}{CB}. On the other side, \angle AOD and \angle AOC together make the straight angle \angle COD, and so do \angle BOD and \angle BOC, so
That makes the two pieces of the major arc on either side of D equal as well.
Finally \triangle OMA is right-angled at M with legs AM and OM = d and hypotenuse OA = r, so the Pythagorean theorem gives AM^2 + d^2 = r^2. This is the same fact the chords lesson reached by measuring along the line; here it comes from congruent triangles.
Result 2(a). Let M be the midpoint of a chord \overline{AB} that is not a diameter, so M \ne O. In \triangle OMA and \triangle OMB: OA = OB (radii), AM = MB (midpoint) and OM = OM, so the triangles are congruent by SSS (the congruence criteria). Hence \angle OMA \cong \angle OMB. These two angles form a linear pair, so they add to 180^\circ, and being equal each is 90^\circ.
Result 2(b). A point lies on the perpendicular bisector of \overline{AB} exactly when it is the same distance from A as from B (the perpendicular bisector theorem). The center is: OA = OB = r. So O is on the perpendicular bisector. Two chords that are not parallel have perpendicular bisectors that are not parallel either, so those two lines cross at exactly one point, and O lies on both: that point is the center.
Result 3. Let O be the center of the circle holding \overline{AB} and O' the center of the one holding \overline{CD} (the same point when there is one circle). All four radii are equal.
- If AB = CD, then \triangle OAB \cong \triangle O'CD by SSS, since OA = O'C, OB = O'D and AB = CD. Matching angles give \angle AOB \cong \angle CO'D. (If the chords are diameters, both angles are straight.)
- If \angle AOB \cong \angle CO'D, then \triangle OAB \cong \triangle O'CD by SAS, with the two pairs of radii as the sides, so AB = CD.
- A minor arc measures the same as its central angle, and two arcs on congruent circles are congruent exactly when their measures are equal. So equal central angles and congruent minor arcs say the same thing.
Result 4. Let the chords have lengths 2a and 2b and distances d and e from the center of circles of the same radius r. By Result 1,
If d = e the right-hand sides match, so a^2 = b^2, and since lengths are positive a = b. If a = b then d^2 = e^2, and since distances are not negative d = e. If d < e then d^2 < e^2, so r^2 - d^2 > r^2 - e^2, so a > b: the nearer chord is the longer.
Worked examples
A chord and its distance from the center
A circle has center O and radius 17. A chord \overline{AB} is 30 long. How far is the chord from the center?
SolutionDrop the perpendicular from O to the chord, meeting it at M. By Result 1, M is the midpoint of \overline{AB}, so AM = 15. The triangle OMA has a right angle at M and hypotenuse OA = 17, a radius, so
so OM = 8. The chord is 8 from the center.
Check: 8^2 + 15^2 = 64 + 225 = 289 = 17^2.
Expressions on the two halves
In a circle with center O, the segment \overline{OM} is perpendicular to the chord \overline{AB} at M, with AM = 2x + 4 and MB = 4x - 8. The radius is 20. Find x, the length of \overline{AB}, and the distance OM.
Solution\overline{OM} lies on a line through the center perpendicular to the chord, so by Result 1 it bisects the chord, and AM = MB:
Then AM = 2(6) + 4 = 16 and MB = 4(6) - 8 = 16, which agree, so AB = 32.
For the distance, \triangle OMA is right-angled at M with hypotenuse OA = 20:
so OM = 12.
Halving the arcs
In a circle with center O, the diameter \overline{CD} is perpendicular to the chord \overline{AB}, with C on the minor arc \overset{\frown}{AB}, and m\overset{\frown}{AB} = 110^\circ. Find m\overset{\frown}{AC}, m\angle AOC, m\overset{\frown}{AD} and the measure of the major arc \overset{\frown}{ADB}.
SolutionBy Result 1 the diameter bisects the minor arc \overset{\frown}{AB} at C, so
\overset{\frown}{AC} is a minor arc, so its central angle has the same measure: m\angle AOC = 55^\circ.
C and D are the ends of a diameter, so \angle AOC and \angle AOD together make a straight angle, and m\angle AOD = 180^\circ - 55^\circ = 125^\circ. That is less than 180^\circ, so \overset{\frown}{AD} is the minor arc with that central angle: m\overset{\frown}{AD} = 125^\circ.
The major arc \overset{\frown}{ADB} runs from A through D to B, and D bisects it, so it is 125^\circ + 125^\circ = 250^\circ. Check: it should be 360^\circ minus the minor arc, and 360^\circ - 110^\circ = 250^\circ.
By Result 3, the equal arcs \overset{\frown}{AC} and \overset{\frown}{CB} also have equal chords, so AC = CB: the point C is the same distance from both ends of the chord, as any point of a perpendicular bisector must be.
The radius of an arch
A stone arch is part of a circle. Its span, the straight distance AB between its feet, is 24 m, and its rise, the height MC from the midpoint M of the span up to the arch at C, is 8 m. Find the radius of the circle.
The segment \overline{MC} is perpendicular to the span at its midpoint, so it lies on the perpendicular bisector of the chord \overline{AB}. By Result 2(b) that line passes through the center O. So O, M and C are on one vertical line.
Call the radius r. Then OC = r and MC = 8, so OM is r - 8 if O is below the span, or 8 - r if it is above; either way OM^2 = (r - 8)^2. The triangle OMA has a right angle at M, legs OM and AM = 12, and hypotenuse OA = r:
The radius is 13 m. Then OM = 13 - 8 = 5, so the center is 5 m below the span.
Check: 5^2 + 12^2 = 25 + 144 = 169 = 13^2.
Two parallel chords
A circle has radius 25. Two parallel chords have lengths 48 and 14. How far apart are the chords?
SolutionThe perpendicular from the center O to one chord is also perpendicular to the other, because the chords are parallel. So one line through O meets both chords at right angles, and by Result 1 it meets each at its midpoint.
For the chord of length 48, the half-chord is 24 and its distance from O is
For the chord of length 14, the half-chord is 7 and its distance is
The problem does not say whether the chords lie on the same side of the center, and both are possible, so there are two answers. On the same side, the chords are 24 - 7 = 17 apart. On opposite sides, they are 24 + 7 = 31 apart.
Result 4 shows in the numbers: the longer chord is the nearer one.
Finding a center from three points
The points A(-3, 1), B(2, 6) and C(5, 5) lie on a circle. Find its center and radius.
Solution\overline{AB} and \overline{BC} are chords, so by Result 2(b) the center lies on the perpendicular bisector of each. Find both lines, then where they cross.
\overline{AB} has midpoint
and slope \displaystyle \frac{6 - 1}{2 - (-3)} = 1. A perpendicular line has slope -1 (slopes of perpendicular lines multiply to -1), so the perpendicular bisector is
\overline{BC} has midpoint \displaystyle \left(\frac{7}{2}, \frac{11}{2}\right) and slope \displaystyle \frac{5 - 6}{5 - 2} = -\frac{1}{3}. A perpendicular line has slope 3, so
Where they cross, -x + 3 = 3x - 5, so 8 = 4x and x = 2, and then y = -2 + 3 = 1. The center is O(2, 1).
The radius is the distance from O to any of the three points. From A, it is \displaystyle \sqrt{(-3 - 2)^2 + (1 - 1)^2}, which is \displaystyle \sqrt{25} = 5.
Check with the other two: \displaystyle OB = \sqrt{0^2 + 5^2} = 5 and \displaystyle OC = \sqrt{3^2 + 4^2} = 5. All three points are 5 from (2, 1), as they must be.
The common chord of two circles
Two circles, with centers P and Q and radii 10 and 17, meet at the points A and B. The centers are 21 apart. Show that \overline{PQ} is perpendicular to the common chord \overline{AB} and passes through its midpoint, then find AB.
P is the center of a circle through A and B, so PA = PB. By the perpendicular bisector theorem, P lies on the perpendicular bisector of \overline{AB}. The same holds for Q, since QA = QB. Two different points fix one line, so the line PQ is the perpendicular bisector of \overline{AB}: it crosses the chord at its midpoint M, at a right angle. This is Result 2(b) for both circles at once.
Put the line PQ on a number line with P at 0 and Q at 21, and let M be at x. Then PM^2 = x^2 and QM^2 = (21 - x)^2, whichever side of P the point M is on. The triangles PMA and QMA both have a right angle at M and share the leg AM, so
Setting the two equal:
So PM = 6 and QM = 15, and M lies between the centers. Then AM^2 = 100 - 36 = 64, so AM = 8 and
Check with the other circle: 15^2 + 8^2 = 225 + 64 = 289 = 17^2.
Boundaries
The line must pass through the center. Many lines are perpendicular to a chord, one through each of its points, and only the one through the midpoint bisects it. A line through the center that is not perpendicular to the chord does not bisect it either, unless the chord is a diameter.
The converse 2(a) needs a chord that is not a diameter. Every line through the center passes through the midpoint of a diameter, which is the center itself, and only one of those lines is perpendicular to it.
Finding the center needs two chords that are not parallel. Parallel chords have the same perpendicular bisector, a single line through the center, and one line does not say where on it the center is.
Results 3 and 4 hold in one circle or in congruent circles. A chord of length 6 in a circle of radius 5 is 4 from its center, while a chord of length 6 in a circle of radius 10 is \displaystyle \sqrt{91} from its center, and the two chords cut arcs of different measures. Equal chords in circles of different sizes match nothing else.
Result 3 pairs a chord with its minor arc. The major arcs of equal chords are also equal, being 360^\circ minus equal minor arcs, but between unequal chords the longer one has the larger minor arc and the smaller major arc.
A perpendicular from the center bisects two arcs, not one. It meets the circle twice, at the middle of the minor arc and at the middle of the major arc. When the chord is a diameter these are the middles of two semicircles.
The distance is the perpendicular distance. The distance from O to an end of the chord is always r and says nothing; only the perpendicular distance d, to the midpoint, decides the length.
Where this goes next
Once a chord is known to be cut in half by the perpendicular from the center, the pieces two crossing chords cut each other into can be compared, which is the product rule in segment lengths in circles. The same symmetry, pushed to the moment a chord shrinks to a single point, is the tangent properties: a tangent is at right angles to the radius where it touches.
Result 2(b), run on three points instead of a circle, is the Circumcenter: the perpendicular bisectors of a triangle’s sides meet at the center of the one circle through all three corners, and constructions with circles draw it with compass and straightedge.
Result 3 explains why equal chords laid end to end around a circle close up into a regular polygon: equal sides cut equal arcs, so each central angle is 360^\circ divided by the number of sides. And the midpoint C of an arc, found by the perpendicular from the center, has a property that the inscribed angle theorem makes precise: since \overset{\frown}{AC} and \overset{\frown}{CB} are equal, the line from any point of the other arc to C splits the inscribed angle at that point into two equal halves.
Exercises
The perpendicular from the center
- A circle has radius 10. A chord is 12 long. How far is it from the center?
- A circle has radius 13. A chord is 5 from the center. How long is the chord?
- A chord 40 long lies 21 from the center of a circle. Find the radius.
- In a circle with center O, the diameter \overline{CD} is perpendicular to the chord \overline{AB} at M, with C on the minor arc. AB = 24 and OM = 9. Find the radius, CM and DM.
- In a circle with center O, \overline{OM} \perp \overline{AB} at M, with AM = 3x - 1 and MB = x + 9. Find x and AB.
Arcs and central angles
- In a circle with center O, the diameter \overline{CD} is perpendicular to the chord \overline{AB}, with C on the minor arc. If m\overset{\frown}{AB} = 128^\circ, find m\overset{\frown}{AC}.
- In the same setting, m\angle AOC = 38^\circ. Find m\overset{\frown}{AB} and the measure of the major arc \overset{\frown}{ADB}.
- In a circle with center O, m\angle AOB = m\angle COD = 72^\circ, AB = 4x + 1 and CD = 6x - 9. Find x and AB.
- In one circle the chords \overline{AB} and \overline{CD} are equally long, with m\overset{\frown}{AB} = (5y + 10)^\circ and m\overset{\frown}{CD} = (7y - 30)^\circ (both minor arcs). Find y and the measure of each arc.
- Five chords of equal length are laid end to end around a circle and close up exactly. What does each of their minor arcs measure?
Distance from the center
- In a circle with center O, \overline{OM} \perp \overline{AB} at M and \overline{ON} \perp \overline{CD} at N, with OM = ON. AB = 7x - 4 and CD = 5x + 8. Find x and AB.
- A circle has radius 17. The chord \overline{AB} is 30 long and the chord \overline{CD} is 16 long. Find each chord’s distance from the center. Which is nearer?
- Two chords of a circle of radius 15 are each 18 long. How far is each from the center?
- In a circle of radius 10, one chord is 6 from the center and another is 8 from it. Find both lengths. Which is longer, and by how much?
Finding centers and radii
- An arc has a chord 16 long, and the arc rises 4 above the chord’s midpoint. Find the radius of its circle.
- The points (0, 0), (8, 0) and (0, 6) lie on a circle. Find its center and radius.
- A circle passes through (-2, 3) and (4, 3), and its center lies on the line y = x. Find the center and the radius.
- A circle has radius 13. Two parallel chords are 24 and 10 long. Find every possible distance between them.
- Two circles of radii 13 and 15 meet, and their common chord is 24 long. Find every possible distance between their centers.
Reasoning
- A line is perpendicular to the chord \overline{AB} but does not pass through the center. Explain why it cannot bisect \overline{AB}.
- Two different chords of a circle bisect each other. Prove that both are diameters.
- Explain, using Result 4, why the longest chord of a circle is a diameter.
- A carpenter wants the center of a round tabletop and has only a ruler and a set square (a tool for drawing right angles). Explain how to find it, and why the two chords used must not be parallel.
- The chords \overline{AB} and \overline{CD} of a circle with center O are equally long and cross at a point P other than O. Prove that \overline{OP} bisects one of the angles between the chords.
Answers
- 8. The perpendicular from the center halves the chord, so the half-chord is 6 and the distance is \displaystyle \sqrt{10^2 - 6^2} = \sqrt{64} = 8.
- 24. The half-chord is \displaystyle \sqrt{13^2 - 5^2} = \sqrt{144} = 12, so the chord is 2 \cdot 12 = 24.
- 29. The half-chord is 20, so \displaystyle r = \sqrt{20^2 + 21^2} = \sqrt{841} = 29.
- Radius 15, CM = 6, DM = 24. M is the midpoint, so AM = 12 and \displaystyle r = \sqrt{12^2 + 9^2}, which is \displaystyle \sqrt{225} = 15. C is on the chord’s side of O, so CM = 15 - 9 = 6; D is on the far side, so DM = 15 + 9 = 24.
- x = 5, AB = 28. The perpendicular from the center bisects the chord, so 3x - 1 = x + 9, 2x = 10, x = 5. Then AM = MB = 14 and AB = 28.
- 64^\circ. The diameter bisects the minor arc, so m\overset{\frown}{AC} = 128^\circ \div 2 = 64^\circ.
- m\overset{\frown}{AB} = 76^\circ and m\overset{\frown}{ADB} = 284^\circ. The minor arc \overset{\frown}{AC} measures 38^\circ, its central angle, and C bisects \overset{\frown}{AB}, so m\overset{\frown}{AB} = 2 \cdot 38^\circ = 76^\circ. The major arc is 360^\circ - 76^\circ = 284^\circ.
- x = 5, AB = 21. Equal central angles give equal chords (Result 3), so 4x + 1 = 6x - 9, 10 = 2x, x = 5, and AB = CD = 21.
- y = 20, each arc 110^\circ. Equal chords cut equal minor arcs (Result 3), so 5y + 10 = 7y - 30, 40 = 2y, y = 20, and 5(20) + 10 = 110.
- 72^\circ. Equal chords cut equal minor arcs (Result 3), and the five arcs fill the circle without overlapping, so each is 360^\circ \div 5 = 72^\circ.
- x = 6, AB = 38. Chords equally far from the center are equally long (Result 4), so 7x - 4 = 5x + 8, 2x = 12, x = 6, and AB = CD = 38.
- \overline{AB} is 8 from the center and \overline{CD} is 15; \overline{AB} is nearer. \displaystyle \sqrt{17^2 - 15^2} = \sqrt{64} = 8 and \displaystyle \sqrt{17^2 - 8^2} = \sqrt{225} = 15. The longer chord is the nearer one, as Result 4 says.
- 12 each. The half-chord is 9, so the distance is \displaystyle \sqrt{15^2 - 9^2} = \sqrt{144} = 12; equal chords are equally far out.
- 16 and 12; the chord 6 from the center is longer, by 4. Half-chords: \displaystyle \sqrt{100 - 36} = 8 and \displaystyle \sqrt{100 - 64} = 6, so the chords are 16 and 12.
- 10. The rise lies on the perpendicular bisector of the chord, which passes through the center. With radius r: (r - 4)^2 + 8^2 = r^2, so -8r + 16 + 64 = 0 and r = 10.
- Center (4, 3), radius 5. The perpendicular bisector of the chord from (0, 0) to (8, 0) is x = 4; of the chord from (0, 0) to (0, 6) it is y = 3. They cross at (4, 3), and the distance to (0, 0) is \displaystyle \sqrt{4^2 + 3^2} = 5.
- Center (1, 1), radius \displaystyle \sqrt{13}. The chord from (-2, 3) to (4, 3) is horizontal with midpoint (1, 3), so its perpendicular bisector is x = 1. The center is on it and on y = x, so it is (1, 1). The radius is \displaystyle \sqrt{(-2 - 1)^2 + (3 - 1)^2}, which is \displaystyle \sqrt{9 + 4} = \sqrt{13}.
- 7 or 17. The chords lie \displaystyle \sqrt{13^2 - 12^2} = 5 and \displaystyle \sqrt{13^2 - 5^2} = 12 from the center, on one line through it. On the same side they are 12 - 5 = 7 apart; on opposite sides 12 + 5 = 17.
- 14 or 4. The line of centers is the perpendicular bisector of the common chord (Example 7), so each center is at a distance from the chord found from the half-chord 12: \displaystyle \sqrt{13^2 - 12^2} = 5 and \displaystyle \sqrt{15^2 - 12^2} = 9. With the centers on opposite sides of the chord they are 5 + 9 = 14 apart; on the same side, 9 - 5 = 4.
- Only one line is perpendicular to \overline{AB} at its midpoint, and by Result 2(b) that line, the perpendicular bisector, passes through the center. A perpendicular that misses the center is therefore a different perpendicular, which crosses the chord’s line somewhere other than the midpoint.
- Let the chords meet at P, the midpoint of each. Suppose P \ne O. Then neither chord is a diameter (a diameter’s midpoint is O), so by Result 2(a) each is perpendicular to \overline{OP} at P. There is only one line through P perpendicular to \overline{OP}, so both chords lie on that line. A line meets a circle in at most two points, so both chords join the same two points and are the same chord, which they are not. So P = O, and both chords pass through the center: both are diameters.
- By Result 4 the nearer a chord is to the center, the longer it is. No distance is smaller than 0, and the chords at distance 0 are the ones through the center, the diameters, of length \displaystyle 2\sqrt{r^2 - 0} = 2r. So a diameter is longer than every chord that misses the center.
- Draw any chord, mark its midpoint by measuring, and use the set square to draw the line through the midpoint at right angles to it. By Result 2(b) the center is on that line. Do the same with a second chord. The center is on both lines, so it is where they cross. If the two chords were parallel, both lines would be the same line through the center, which pins the center to a line but not to a point.
- Let M and N be the feet of the perpendiculars from O to \overline{AB} and \overline{CD}. The chords are equal, so OM = ON by Result 4. Neither foot is P: if M = P, then ON = OM = OP, but in a right triangle the leg ON is shorter than the hypotenuse OP unless N = P too, and then both chords would be perpendicular to \overline{OP} at P and lie on one line, making them the same chord. So \triangle OMP and \triangle ONP are right triangles with the same hypotenuse \overline{OP} and equal legs OM = ON, so they are congruent by hypotenuse leg, and \angle OPM \cong \angle OPN. The rays PM and PN run along the two chords, so \overline{OP} bisects the angle between those rays.