Circle

Work with the friend next to you. Look at the three circles below, each with its center O, and name every part of a circle you can find in them: which line, which curve, which shaded region. Then compare your list with the table.

Part	What it is	In the figures
center	the point inside the circle the same distance from every point on it	O
radius	a line from the center to the circle (and its length, r)	OE, OA, OB, OP
circumference	the circle itself, the curved line all round (and its length, $2\pi r$)	the whole round line
chord	a straight line joining two points of the circle	XY, CD, XZ, and AB too
diameter	a chord through the center; it is two radii, 2r	AB
arc	a piece of the circumference between two of its points	the arc PB, the arc XZ
semicircle	half the circle, the arc cut off by a diameter	each half of the first circle, on either side of AB
sector	the region between two radii and the arc they cut off, a slice of the circle	the shaded slice OPB
segment	the region between a chord and its arc	the shaded piece beside XZ

A chord is a straight line joining two points on a circle. A diameter is a chord too, the one that passes through the center, and it is the longest chord a circle has.

Now a fold. Draw a circle with center O on tracing paper or thin plain paper, and draw any chord AB. Fold the paper along a line through O so that the two halves of the circle lie exactly on each other and A lands on B. Open it out and measure: where does the fold cut AB, and at what angle?

The fold cuts AB into two equal parts, and at a right angle. That is no accident. Folding along a line through the center lays the circle on itself, so it is a line of symmetry of the circle; when A lands on B, each half of the chord lands on the other half. The four theorems of this unit are this fold, measured and then proved.

Theorem 1

The perpendicular drawn from the center of a circle to a chord bisects the chord.

Measuring checks the theorem for the circles you drew. A proof shows it for every circle there could be.

Given: O is the center of the circle, and AB is a chord with $OC \perp AB$.

To prove: AC = BC.

Construction: join O to A and to B.

	Statements	Reasons
1.	In △OCA and △OCB: i. ∠OCA = ∠OCB (R) ii. OA = OB (H) iii. OC = OC (S)	i.both 90∘°, given $OC \perp AB$ ii. radii of the same circle iii. common side
2.	$\triangle OCA \cong \triangle OCB$	by R.H.S.
3.	AC = BC	corresponding sides of congruent triangles

Proved. The two radii make △OAB isosceles, and the perpendicular from its vertex O halves its base: it is the isosceles triangle of Unit 13 again.

Theorem 2

The line joining the center of a circle and the midpoint of a chord is perpendicular to the chord.

This is Theorem 1 turned round, its converse: Theorem 1 starts from a right angle and finds equal halves; Theorem 2 starts from equal halves and finds a right angle.

Given: O is the center of the circle, and M is the midpoint of the chord CD: CM = MD.

To prove: $OM \perp CD$.

Construction: join O to C and to D.

	Statements	Reasons
1.	In △OMC and △OMD: i. OC = OD (S) ii. OM = OM (S) iii. CM = DM (S)	i.radii of the same circle ii. common side iii. given
2.	$\triangle OMC \cong \triangle OMD$	by S.S.S.
3.	∠OMC = ∠OMD	corresponding angles of congruent triangles
4.	$\angle OMC + \angle OMD = 180^\circ$, so $2\angle OMC = 180^\circ$ and $\angle OMC = 90^\circ$	adjacent angles on the straight line CD, and 3
5.	$OM \perp CD$	from 4

Proved.

The activity shows one more thing worth keeping: the perpendicular bisector of every chord passes through the center. (By Theorem 2, the line from O to the chord’s midpoint is perpendicular to it; and there is only one line through the midpoint at right angles to the chord.) That is how the opening widget found the pot’s center: the perpendicular bisectors of two chords each pass through it, so they cross there.

Theorem 3

Equal chords of a circle are equidistant from the center.

The distance of a chord from the center is the length of the perpendicular from the center to it: the shortest way from O to the chord.

The verification is all your course asks for here. But the proof is short, and it uses only Theorem 1 and R.H.S., so here it is as an extra.

Given: AB = CD, $OM \perp AB$ and $ON \perp CD$.

To prove: OM = ON.

Construction: join OA and OC.

	Statements	Reasons
1.	AM is half of AB and CN half of CD, so AM = CN	perpendiculars from the center bisect the chords (Theorem 1); AB = CD
2.	In △OMA and △ONC: i. ∠OMA = ∠ONC (R) ii. OA = OC (H) iii. AM = CN (S)	i.both 90∘° ii. radii iii. from 1
3.	$\triangle OMA \cong \triangle ONC$	by R.H.S.
4.	OM = ON	corresponding sides of congruent triangles

Proved.

Theorem 4 (converse of Theorem 3)

Chords which are equidistant from the center of a circle are equal.

Again an extra proof, the same triangles the other way round.

Given: $OM \perp AB$, $ON \perp CD$ and OM = ON.

To prove: AB = CD.

	Statements	Reasons
1.	In △OMA and △ONC: i. ∠OMA = ∠ONC (R) ii. OA = OC (H) iii. OM = ON (S)	i.both 90∘° ii. radii iii. given
2.	$\triangle OMA \cong \triangle ONC$	by R.H.S.
3.	AM = CN	corresponding sides of congruent triangles
4.	AB = 2AM = 2CN = CD	perpendiculars from the center bisect the chords (Theorem 1), and 3

Proved. Both theorems come from one picture: the radius, the distance and half the chord make a right-angled triangle. Fix the radius and one of the other two, and the third is fixed too. Try it:

Worked examples

Most calculations in this unit are one right-angled triangle: the radius r as its hypotenuse, the distance d of the chord from the center, and half the chord, c/2:
$$r^2 = d^2 + \left(\frac{c}{2}\right)^2$$
Draw that triangle first, and the rest is Pythagoras’ rule.

Back to the broken pot

Sapana lays a ruler across the inside of her piece of rim, touching its curve at both ends: the ruler is a chord, AB = 20 cm. At the middle of the ruler she measures straight across to the curve: MS = 4 cm. (The middle of an arc’s chord and the middle of the arc are on the same line through the center, at right angles to the chord.)

Let the pot’s radius be r. The center O is on the line through M and S, so OS = r and OM = r − 4. In the right-angled △OMA, with AM = 10 cm:
$$\begin{aligned} r^2 &= (r - 4)^2 + 10^2 \\ r^2 &= r^2 - 8r + 16 + 100 \\ 8r &= 116 \\ r &= 14.5 \end{aligned}$$

The pot’s rim had a radius of 14.5 cm: the pot was 29 cm across at the mouth, a wide pot. Two measurements of a small piece gave the whole circle.

The same steps work for any chord c and height h (the sagitta, Latin for “arrow”, since the chord and arc look like a bow with its arrow):
$$r = \dfrac{h^2 + \left(\dfrac{c}{2}\right)^2}{2h}$$

The opening widget used three points on a piece of rim. Turn it round: given any three points A, B and C not on one straight line, is there a circle through all three?

The center would have to be on the perpendicular bisector of AB (since AB would be a chord), and on that of BC as well. Two lines that aren’t parallel cross at exactly one point, O. Every point on the bisector of AB is equally far from A and B, and every point on the bisector of BC equally far from B and C, so OA = OB = OC: the circle with center O and radius OA passes through all three. So through any three points not on a line there is exactly one circle.

For a triangle ABC, this is its circumcircle, the circle through its three corners, and O its circumcenter. The third bisector, of CA, passes through O too, since OC = OA. (The circle inside a triangle, touching its three sides, the incircle, is found the other way: its center is where the bisectors of the triangle’s angles meet.)

Grade 9’s final exam is set to CDC’s specification grid for Compulsory Mathematics, Grades 9–10. There, Geometry is 13 marks in three questions; in the grid’s model paper, this unit is question 13:

Part	Level	Marks	What it asks
a.	knowledge	1	what kind of triangle the two radii and the chord make
b.	understanding	2	the radius from the chord and its distance (Pythagoras)
c.	higher ability	1	whether a circle of a given radius can pass through the chord’s ends, with a reason

Here is one of our own in that shape.

Question 1

Question 2

Question 3

Question 4