Circle

In Grade 9 you met three facts about chords. Each can be checked with a ruler, a set square and a pair of compasses: that is, verified experimentally.

a. The perpendicular drawn from the center of a circle to a chord bisects the chord: in Figure 12.1, $OM \perp AB$, and AM = MB.

b. The other way round: the line joining the center to the midpoint of a chord is perpendicular to the chord.

c. Chords that are the same distance from the center are equal.

To verify one, draw it in two or three circles of different sizes, measure, and put the numbers in a table. For a, measure AM and MB in each circle; for c, draw two chords with OM = ON (as in Figure 12.2) and measure AB and CD. If the two columns agree in every circle, to within a millimeter or so of measuring error, the statement holds for the circles you drew. You will write a verification like this in full later in the unit.

• An angle whose vertex is the center of a circle and whose arms are two radii is a central angle (∠AOB in Figure 12.3).
• An angle whose vertex is on the circle and whose arms are two chords is an inscribed angle, also called an angle at the circumference or a circumference angle (∠PQR in Figure 12.4).
• Two points on a circle cut it into two arcs. The shorter is the minor arc, the longer the major arc. A minor arc is named by its two ends (arc AB); a major arc by its ends and a point on it (arc APB), so there is no doubt which one is meant.

An inscribed angle stands on (is based on) the arc between its arms, on the far side from its vertex: ∠APB in Figure 12.5 stands on the minor arc AB, and so does the central angle ∠AOB.

How big is an arc? Its length depends on the size of the circle, but its share of the circle does not. A quarter of a small circle and a quarter of a big one are both a quarter.

So an arc can be measured by its central angle. The full circle is 360∘°, and an arc is given the degree measure of its central angle: arc AC in Figure 12.8 is a 60∘° arc. This is written
$$\angle AOC \;≙\; \text{arc } AC,$$
read “the angle AOC corresponds to the arc AC” (the angle and the arc have the same number of degrees).

Now the angle at the circumference. In the figure, ∠ACB stands on the arc AB, and OC is joined.

Follow the angles. In the isosceles △OAC, the angles add up to 180∘°, and the two base angles are equal:
$$\begin{aligned} 2\angle OCA &= 180° - \angle AOC \quad \text{(i)} \end{aligned}$$
In the same way, in △OBC:
$$\begin{aligned} 2\angle OCB &= 180° - \angle BOC \quad \text{(ii)} \end{aligned}$$
Adding (i) and (ii):
$$\begin{aligned} 2(\angle OCA + \angle OCB) &= 360° - (\angle AOC + \angle BOC) \\ 2\angle ACB &= 360° - \text{reflex } \angle AOB \\ 2\angle ACB &= \angle AOB \end{aligned}$$
The last line holds because the reflex angle AOB and the angle AOB make a full turn, 360∘°. So
$$2\angle ACB \;≙\; \text{arc } AB.$$

An inscribed angle is half the degree measure of the arc it stands on. In degrees, the arc, the central angle on it and twice the inscribed angle on it are all the same number.

A. On the same arc

Paper folding shows it once. Experimental verification shows it with measurements, in a form the SEE marks. Draw at least two circles (the SEE asks for a radius of at least 3 cm) and, in each, an inscribed angle and a central angle standing on the same arc.

Measure each angle with a protractor and fill in a table:

Fig.	∠AOB	∠ACB	Result
1	130∘°	65∘°	∠AOB = 2∠ACB
2	50∘°	25∘°	∠AOB = 2∠ACB
3	240∘° (reflex)	120∘°	∠AOB = 2∠ACB

Conclusion: the central angle is twice the inscribed angle standing on the same arc.

Given: O is the center of the circle. The central angle AOB and the inscribed angle ACB stand on the same arc AB.

To prove: ∠AOB = 2∠ACB.

Construction: join C to O and produce CO to D.

The idea: CO cuts the picture into two isosceles triangles, and each has an exterior angle at O.

	Statements	Reasons
1.	In △AOC: i. ∠OAC = ∠OCA ii. ∠AOD = ∠OAC + ∠OCA iii. ∠AOD = 2∠OCA	i.OA = OC (radii), base angles of an isosceles triangle ii. an exterior angle equals the sum of the two opposite interior angles iii. from i and ii
2.	In △BOC: ∠BOD = 2∠OCB	as in 1, since OB = OC
3.	∠AOD + ∠BOD = 2∠OCA + 2∠OCB	adding 1 iii and 2
4.	∠AOB = 2∠ACB	from 3, the whole is the sum of its parts

Proved.

The angle in a semicircle

Make AB a diameter. The central angle AOB is now a straight angle, 180∘°, so the inscribed angle on it is half of 180∘°.

B. Inscribed angles on the same arc

Given: O is the center of a circle, and the inscribed angles ACB and ADB stand on the same arc APB.

To prove: ∠ACB = ∠ADB.

Construction: join O to A and to B.

	Statements	Reasons
1.	∠AOB = 2∠ACB	the central angle is double the inscribed angle on the same arc APB (Theorem 1)
2.	∠AOB = 2∠ADB	the same reason
3.	2∠ACB = 2∠ADB	from 1 and 2
4.	∠ACB = ∠ADB	halving both sides of 3

Proved.

So all the inscribed angles on one arc are equal: the points C and D can be anywhere on the other arc. That is the curve in the opening: every spot on the arc sees the row at the same angle.

A quadrilateral whose four corners all lie on one circle is a cyclic quadrilateral.

Given: O is the center of the circle and ABCD is a cyclic quadrilateral.

To prove: $\angle ABC + \angle ADC = 180°$ and $\angle BCD + \angle BAD = 180°$.

Construction: join O to A and to C.

The idea: ∠ADC and ∠ABC stand on the two arcs AC, which together make the whole circle.

	Statements	Reasons
1.	obtuse ∠AOC = 2∠ADC	the central angle and the inscribed angle on the same arc ABC
2.	reflex ∠AOC = 2∠ABC	the central angle and the inscribed angle on the same arc ADC
3.	2∠ADC + 2∠ABC = obtuse ∠AOC + reflex $\angle AOC = 360°$	from 1 and 2; the angles round the point O make 360∘°
4.	$\angle ADC + \angle ABC = 180°$	halving 3
5.	$\angle BCD + \angle BAD = 180°$	in the same way, joining OB and OD (or: the four angles of a quadrilateral make 360∘°)

Proved.

The converse is true too, and the exercise needs it: if a pair of opposite angles of a quadrilateral add up to 180∘°, its four corners lie on one circle, so it is a cyclic quadrilateral. And since an exterior angle and its interior angle make 180∘°, a quadrilateral is cyclic when an exterior angle equals the interior opposite angle (Example 5 shows why these go together).

Worked examples

Back to the photo

The photographer can stand anywhere on an arc of a circle through the two ends of the row, on the yard’s side: every point of that arc sees the row at the same 60∘°, because all those angles stand on the same chord, the row (Theorem 2).

Which circle? The angle at the center is twice 60∘°, that is 120∘°. The center is on the perpendicular bisector of the row, and the triangle it makes with the row’s ends has a 120∘° angle and two 30∘° angles, so the center is $6 \times \tan 30° \approx 3.46$ m from the row and the radius is $6 \div \sin 60° \approx 6.93$ m. The farthest spot, straight in front of the middle, is $3.46 + 6.93 \approx 10.4$ m from the row.

In the SEE’s Compulsory Mathematics paper, Geometry carries 13 marks in three questions (CDC’s specification grid). The question on circles is worth 4 marks:

Part	Level	Marks	What it asks
a.	knowledge	1	state a relation between angles in a circle
b.	understanding	1	find an unknown angle or a value of x
c.	application	2	verify a relation by measuring, in at least two circles of radius 3 cm or more

Here is a question in that shape.

Question 1

If O is the center of each circle, find the value of x.

Question 2

Find the value of x.

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

These questions mix Units 10, 11 and 12, as the SEE’s geometry questions do.

Question 1

ABCD is a parallelogram and ABFE is a rectangle.

Question 2

A parallelogram ABCD and a rectangle ABFE stand on the same base AB between the same parallel lines AB and EC.

Question 3

A parallelogram BCED and a triangle ABC stand on the same base BC between the same parallels BC and DE, with $\angle BAC = 90°$, AB = 7 cm and BC = 25 cm.

Question 4

ABCD and PQRD are parallelograms.

Question 5

ABCD is a parallelogram and X is a point inside it, with $MN \parallel AB$ through X.

Question 6

Question 7

Question 8

A circle with center O has a central angle BOC and an inscribed angle BDC on the same arc BC.

Question 9

ABCD is a cyclic quadrilateral.

Question 10

P, Q, R and S are points on a circle, and AP = AQ.

Question 11

O is the center of the circle and SH a diameter. S, I, H and U are on the circle, L is outside it, and $LO \perp SH$.

Question 12

O is the center of the circle, ∠PQS = y, $\angle QSR = 63°$ and ∠PRQ = x.