Chapter 5: I'm Up and Down, and Round and Round (Circles)

Steps of construction:

• Draw AB = 5 cm. At A draw an angle of 70° and at B an angle of 60°. The rays meet at C (then ∠C = 180° − 70° − 60° = 50°).
• Draw the perpendicular bisectors of AB and AC. They meet at O.
• With centre O and radius OA, draw the circle.

All three angles (70°, 60°, 50°) are less than 90°, so the triangle is acute-angled, and the circumcentre O lies inside it. (By calculation, the radius is about 3.26 cm.)

The circumcentre lies inside the triangle (the triangle is acute-angled).

Steps of construction:

• Draw AB = 5 cm. At A draw an angle of 100° and cut AC = 4 cm on its arm. Join BC.
• Draw the perpendicular bisectors of AB and AC; they meet at O.
• With centre O and radius OA, draw the circle.

∠A = 100° is obtuse, so the circumcentre lies outside the triangle, on the other side of the longest side BC. (Radius ≈ 3.5 cm.)

The circumcentre lies outside the triangle (because ∠A is obtuse).

Steps of construction:

• Draw AB = 6 cm. With centres A and B and radius 7 cm, draw arcs meeting at C. Join CA and CB.
• Draw the perpendicular bisectors of AB and BC; they meet at O.
• With centre O and radius OA, draw the circle.

On measuring, $OA=OB=OC\approx 3.9$ cm. They are equal because O is on the perpendicular bisector of AB (so OA = OB) and on that of BC (so OB = OC). By calculation:

$$\text{Height from C}=\sqrt{7^2-3^2}=\sqrt{40}\approx 6.32\text{cm},R=\frac{abc}{4\times \text{Area}}=\frac{6\times 7\times 7}{4\times \frac{1}{2}\times 6\times 6.32}\approx 3.87\text{cm}$$

OA = OB = OC ≈ 3.9 cm (the circumradius).

The centre of any circle through A and B is equidistant from A and B, so it lies on the perpendicular bisector of AB. The radius is the distance from the centre O to A. Of all points on the perpendicular bisector, the one closest to A is the midpoint M of AB (for any other point O, OA is the hypotenuse of right triangle OMA, so $OA>MA$).

So the smallest circle has its centre at the midpoint of AB and AB as a diameter.

The least radius is $\frac{1}{2}AB$ (the circle with AB as diameter).

Let AB be a chord of a circle with centre O. In △OAB, OA and OB are both radii, so $OA=OB$. A triangle with two equal sides is isosceles. (Also ∠OAB = ∠OBA, the base angles.)

If the chord is a diameter, O lies on AB and there is no triangle; for every other chord the triangle exists and is isosceles.

OA = OB = radius, so △OAB is isosceles.

Let AB and CD be chords of the same circle (centre O) with $AB=CD$. In △OAB and △OCD:

• $OA=OC$ (radii)
• $OB=OD$ (radii)
• $AB=CD$ (given)

So $△OAB\cong △OCD$ by the SSS congruence rule.

By SSS (OA = OC, OB = OD, AB = CD), the two triangles are congruent.

Let C be the centre and CM ⊥ AB, with M on the chord AB. In △CMA and △CMB:

• $\angle CMA=\angle CMB=90°$
• $CA=CB$ (radii, the hypotenuses)
• $CM=CM$ (common side)

So $△CMA\cong △CMB$ by the RHS congruence rule. Hence $AM=BM$, i.e. M is the midpoint of AB.

By RHS congruence of △CMA and △CMB, AM = BM, so the perpendicular from the centre bisects the chord.

Let AD be the altitude from A, so AD ⊥ BC. In △ABD and △ACD: $AB=AC$ (given), $\angle ADB=\angle ADC=90°$, and AD is common. By RHS, $△ABD\cong △ACD$, so $BD=DC$.

So AD is perpendicular to BC and passes through its midpoint D: AD lies along the perpendicular bisector of the chord BC. The perpendicular bisector of a chord passes through the centre (the centre O is equidistant from B and C). Hence the altitude AD passes through O.

The altitude from A bisects BC at right angles, so it is the perpendicular bisector of chord BC, which always passes through the centre.

The perpendicular from the centre O bisects each chord. Because the chords are parallel, both perpendiculars lie on one line through O, and the midpoints M and N lie on it.

For the 8 cm chord PQ: $MQ=4$, so $OM=\sqrt{5^2-4^2}=\sqrt{9}=3$ cm.

For the 6 cm chord RS: $NS=3$, so $ON=\sqrt{5^2-3^2}=\sqrt{16}=4$ cm.

The chords are on opposite sides of O, so

$$MN=OM+ON=3+4=7\text{cm}$$

The distance between the midpoints is 7 cm.

Theorem 6: equal chords are equidistant from the centre. Let AB and FG be equal chords of a circle with centre C and radius r, and let E, H be the feet of the perpendiculars from C (they are the midpoints).

In right △CEA: $C{E}^2=C{A}^2-A{E}^2=r^2-{\left(\frac{AB}{2}\right)}^2$.

In right △CHF: $C{H}^2=C{F}^2-F{H}^2=r^2-{\left(\frac{FG}{2}\right)}^2$.

Since $AB=FG$, the right-hand sides are equal, so $C{E}^2=C{H}^2$ and hence $CE=CH$.

$C{E}^2=r^2-\frac{A{B}^2}{4}$ and $C{H}^2=r^2-\frac{F{G}^2}{4}$; with AB = FG these are equal, so CE = CH.

E and H are the midpoints of AB and GF (the perpendicular from the centre bisects a chord). In △CEA and △CHF:

• $\angle CEA=\angle CHF=90°$
• $CA=CF$ (radii, the hypotenuses)
• $CE=CH$ (given)

So $△CEA\cong △CHF$ (RHS). Hence $AE=FH$, and doubling, $AB=2AE=2FH=GF$.

By RHS congruence, AE = FH; so AB = 2AE = 2FH = GF.

In right △CEA: $A{E}^2=C{A}^2-C{E}^2=r^2-C{E}^2$. In right △CHF: $F{H}^2=r^2-C{H}^2$. Since $CE=CH$, we get $A{E}^2=F{H}^2$, so $AE=FH$. As E and H are midpoints, $AB=2AE=2FH=GF$.

$A{E}^2=r^2-C{E}^2=r^2-C{H}^2=F{H}^2$, so AE = FH and AB = GF.

In-text exercise (Fig. 5.19): measuring with a protractor, arc AKB subtends about 100° at O (less than 180°), so it is a minor arc; arc CLD subtends about 205° at O (more than 180°), so it is a major arc. (Your measured values may differ slightly.)

The perpendicular from the centre bisects the chord. If half the chord is x, then by the Baudhayana-Pythagoras theorem:

$$x=\sqrt{7^2-6^2}=\sqrt{49-36}=\sqrt{13}$$

$$\text{Chord}=2\sqrt{13}\approx 7.21\text{cm}$$

The chord is $2\sqrt{13}$ cm ≈ 7.21 cm long.

Let AB be the chord, O the centre and OM ⊥ AB with OM = d. Then M is the midpoint of AB (Theorem 5) and △OMA is right-angled at M with hypotenuse OA = r.

$$A{M}^2=O{A}^2-O{M}^2=r^2-d^2\Rightarrow AM=\sqrt{r^2-d^2}$$

$$AB=2AM=2\sqrt{r^2-d^2}$$

Half the chord is $\sqrt{r^2-d^2}$ (Pythagoras in △OMA), so the chord is $2\sqrt{r^2-d^2}$.

No. Let CD be at distance d and AB at distance 2d. Then

$$CD=2\sqrt{r^2-d^2},AB=2\sqrt{r^2-4d^2}$$

The chord length does not change in proportion to the distance. Counter-example: r = 5 cm, d = 2 cm. Then $CD=2\sqrt{21}\approx 9.17$ cm and $AB=2\sqrt{25-16}=6$ cm. But $2AB=12$ cm ≠ 9.17 cm. (Indeed CD can never exceed the diameter 10 cm, while 2AB could.) All we can say is that CD is longer than AB, since it is closer to the centre.

No. For example, with r = 5 and distances 2 and 4: CD ≈ 9.17 cm, AB = 6 cm, and 9.17 ≠ 12. We only know CD > AB.

In △OAB, OA = OB = 12 cm, so it is isosceles and its base angles are equal: $\angle OAB=\angle OBA=\frac{180°-60°}{2}=60°$. All three angles are 60°, so △OAB is equilateral, and AB = OA.

AB = 12 cm.

(i) No. X and Y on the same side of AB lie on the same arc, and angles in the same segment are equal (each is half the central angle of the other arc).

(ii) Not always. If X is on one side and Y on the other, then $\angle AXB+\angle AYB=180°$ (they are opposite angles of the cyclic quadrilateral AXBY). They can be equal only when both are 90°, i.e. when AB is a diameter. So: if AB is not a diameter, equal angles force X and Y to be on the same side; if AB is a diameter, every point of the circle gives 90°, on either side.

(iii) Yes, provided X and Y are on the same side of AB: by Theorem 10, then A, B, X, Y are concyclic, so the circle through A, B, X passes through Y. If X and Y are on opposite sides of AB, this need not be true.

(i) No (angles in the same segment are equal). (ii) Yes, unless AB is a diameter (then both angles are 90° on either side). (iii) Yes, if X and Y are on the same side of AB (Theorem 10).

In Fig. 5.26, ABCD is a quadrilateral inscribed in the circle, with ∠D = 100° and ∠B = x. D and B are opposite vertices.

Opposite angles of a cyclic quadrilateral add up to 180° (Theorem 11):

$$x+100°=180°\Rightarrow x=80°$$

x = 80°

In-text exercise (after Theorem 11): A cyclic quadrilateral with ∠A = 80°, ∠B = 110°, ∠C = 100°, ∠D = 70°. Here ∠A + ∠C = 180° and ∠B + ∠D = 180° (and the total is 360°), so by Theorem 12 such a quadrilateral can be drawn and it will be cyclic.

$$\text{Chord}=2\sqrt{r^2-d^2}=2\sqrt{13^2-5^2}=2\sqrt{169-25}=2\sqrt{144}=2\times 12=24\text{cm}$$

24 cm

By Theorem 9, the angle at a point on the remaining part of the circle is half the angle at the centre:

$$\frac{70°}{2}=35°$$

(If the point is on the arc itself, the angle is $180°-35°=145°$, since it is then the opposite angle of a cyclic quadrilateral.)

35° (at any point of the circle outside the arc).

Radius r = 13 cm, half-chord = 12 cm.

$$d=\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\text{cm}$$

5 cm

$$\text{Chord}=2\sqrt{15^2-9^2}=2\sqrt{225-81}=2\sqrt{144}=24\text{cm}$$

24 cm

Let AB be a chord of a circle with centre O. Every point on the perpendicular bisector of AB is equidistant from A and B, and every point equidistant from A and B lies on it. Since $OA=OB$ (radii), O is equidistant from A and B. Hence O lies on the perpendicular bisector of AB.

Another way: let M be the midpoint of AB. By Theorem 4, OM ⊥ AB. So the line OM is perpendicular to AB at its midpoint, that is, OM is the perpendicular bisector of AB, and it passes through O.

The centre O is equidistant from the ends of the chord (OA = OB), so it lies on the perpendicular bisector of the chord.

The arc AB not containing C subtends a straight angle, ∠AOB = 180°, at the centre. By Theorem 9, the angle at C is half of this.

$$\angle ACB=\frac{1}{2}\times 180°=90°$$

∠ACB = 90° (the angle in a semicircle is a right angle).

Opposite angles of a cyclic quadrilateral are supplementary.

$$\angle C=180°-75°=105°,\angle D=180°-110°=70°$$

∠C = 105° and ∠D = 70°.

P and R are opposite vertices, so

$$(2x+10)+(3x-20)=180\Rightarrow 5x-10=180\Rightarrow x=38$$

$$\angle P=2(38)+10=86°,\angle R=3(38)-20=94°$$

Check: 86° + 94° = 180° ✓

x = 38, ∠P = 86°, ∠R = 94°.

Half the chord is 8 cm. In the right triangle formed by the radius, the half-chord and the perpendicular:

$$r=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\text{cm}$$

The radius is 10 cm.

Case 1: equal sides adjacent (a kite). Let AB = AD = 5 and CB = CD = 12. Then △ABC ≅ △ADC (SSS, AC common), so ∠B = ∠D. Being opposite angles of a cyclic quadrilateral, ∠B + ∠D = 180°, so ∠B = ∠D = 90°.

$$\text{Area}=2\times \text{area of }△ABC=2\times \frac{1}{2}\times 5\times 12=60\text{square units}$$

Case 2: equal sides opposite (5, 12, 5, 12 in order). Then both pairs of opposite sides are equal, so it is a parallelogram. A cyclic parallelogram is a rectangle (see Q14), so the area is $5\times 12=60$ square units.

The area is 60 square units (in either arrangement).

Key fact: for a chord XY and a point Z on the circle, the centre lies on the same side of XY as Z exactly when ∠XZY is acute; it lies on XY when ∠XZY = 90°; and it lies on the opposite side when ∠XZY is obtuse.

Method: for each side of the quadrilateral ABCD, look at the angle it subtends at one of the two other vertices, using a diagonal. For example, side AB subtends ∠ACB; side BC subtends ∠BDC; side CD subtends ∠CAD; side DA subtends ∠DBA.

• If all four of these angles are acute, the centre is inside the quadrilateral.
• If one of them is 90°, the centre lies on that side (that side is a diameter).
• If one of them is obtuse, the centre lies outside, beyond that side.

Only the longest side can subtend an obtuse angle, so in practice the best way is to check the longest side: measure the angle it subtends at an opposite vertex (formed with a diagonal). Acute means inside, right means on the side, obtuse means outside.

Check the angle that each side (in practice, the longest side) subtends at a vertex not on it: all acute means the centre is inside; a right angle means it is on that side; an obtuse angle means it is outside.

Let equal chords AB and CD meet at P, and let O be the centre. Draw OM ⊥ AB and ON ⊥ CD. Then M, N are the midpoints of AB and CD, and since equal chords are equidistant from the centre, $OM=ON$.

In right △OMP and right △ONP: hypotenuse OP is common and $OM=ON$. So $△OMP\cong △ONP$ (RHS) and $MP=NP$.

Also $AM=\frac{1}{2}AB=\frac{1}{2}CD=CN$. Labelling A and C as the ends on the same side as the matching midpoint shifts, we get

$$AP=AM+MP=CN+NP=CP,PB=AB-AP=CD-CP=PD$$

(If P lies between A and M instead, subtract: AP = AM − MP = CN − NP = CP.) So the parts of one chord equal the corresponding parts of the other.

With OM = ON (equal chords) and OP common, △OMP ≅ △ONP, so MP = NP; adding to or subtracting from the equal halves gives AP = CP and PB = PD.

Calculation: half-chord = 3 cm and distance = 3 cm, so the radius is $\sqrt{3^2+3^2}=3\sqrt{2}\approx 4.24$ cm. (△OAB is a right isosceles triangle with ∠AOB = 90°, so the circle is the circumcircle of △OAB.)

Steps of construction:

• Draw AB = 6 cm.
• Draw the perpendicular bisector of AB; let it meet AB at M.
• On the bisector, mark O with OM = 3 cm.
• With centre O and radius OA, draw the circle. It passes through A and B, and AB is at distance OM = 3 cm from O.

Radius = 3√2 ≈ 4.24 cm; the circle is centred on the perpendicular bisector of AB, 3 cm from AB.

Let ABCD be a parallelogram inscribed in a circle.

• In a parallelogram, opposite angles are equal: $\angle A=\angle C$.
• In a cyclic quadrilateral, opposite angles are supplementary: $\angle A+\angle C=180°$.

So $2\angle A=180°$, i.e. $\angle A=\angle C=90°$; similarly $\angle B=\angle D=90°$. A parallelogram with all angles 90° is a rectangle. (Conversely, every rectangle is cyclic, since its opposite angles add to 180°.)

Opposite angles must be both equal (parallelogram) and supplementary (cyclic), so each is 90°: the parallelogram is a rectangle.

Let ABCD be the rectangle. Since $\angle ABC=90°$, the chord AC subtends a right angle at B on the circle. An inscribed right angle is half of a straight angle at the centre, so AC subtends 180° at the centre: AC is a diameter and passes through the centre O, with O as its midpoint. Similarly, $\angle BCD=90°$ makes BD a diameter, with midpoint O.

The diagonals of a rectangle bisect each other, so they meet at their common midpoint, which is O.

Each diagonal subtends 90° at a vertex, so each is a diameter; both diameters have the centre as midpoint, so the diagonals meet at the centre.

Let the circle have centre O and radius r, and let every chord have length l. For any such chord with midpoint M, OM is perpendicular to the chord, so

$$OM=\sqrt{r^2-{\left(\frac{l}{2}\right)}^2}$$

which is the same number for every chord. So all the midpoints are at a fixed distance from O, and as the chord turns round, the midpoint goes all the way round.

A circle with the same centre and radius $\sqrt{r^2-\frac{l^2}{4}}$ (a single point, the centre, when the chords are diameters).

Join OA, OB and OC. In △OAB and △OAC:

• $OB=OC$ (radii)
• $AB=AC$ (given)
• $OA=OA$ (common)

So $△OAB\cong △OAC$ (SSS). Hence $\angle OAB=\angle OAC$: the line AO divides ∠BAC into two equal angles, i.e. O lies on the bisector of ∠BAC.

By SSS, △OAB ≅ △OAC, so ∠OAB = ∠OAC and O is on the bisector of ∠BAC.

Let the radius be r. The distances of the chords from the centre are

$$d_1=\sqrt{r^2-5^2}(\text{10 cm chord}),d_2=\sqrt{r^2-12^2}(\text{24 cm chord})$$

The shorter chord is farther from the centre, and both are on the same side, so $d_1-d_2=7$:

$$\sqrt{r^2-25}=7+\sqrt{r^2-144}$$

Squaring both sides:

$$r^2-25=49+14\sqrt{r^2-144}+r^2-144\Rightarrow 70=14\sqrt{r^2-144}\Rightarrow \sqrt{r^2-144}=5$$

So $r^2=144+25=169$ and $r=13$ cm. Check: $d_1=\sqrt{169-25}=12$, $d_2=5$, and $12-5=7$ ✓

The radius is 13 cm.

Join the centre O to all six vertices. Each side subtends $\frac{360°}{6}=60°$ at O. In △OAB, OA = OB = r and ∠AOB = 60°, so the base angles are also 60° and the triangle is equilateral. So each side of the hexagon is r.

The distance of side AB from O is the altitude OM of the equilateral triangle (M is the midpoint, AM = r/2):

$$OM=\sqrt{r^2-{\left(\frac{r}{2}\right)}^2}=\sqrt{\frac{3r^2}{4}}=\frac{\sqrt{3}}{2}r$$

Each side = r; distance of each side from the centre = $\frac{\sqrt{3}}{2}r$.

The vertices are in the order M, N, O, P around the circle.

• ∠MOP and ∠MNP are both angles subtended by the chord MP at points (O and N) on the same arc of the circle (the arc from M to P through N and O). Angles in the same segment are equal, so $\angle MOP=\angle MNP$.
• Since MN is a diameter, the angles it subtends at O and P are right angles: $\angle MON=\angle MPN=90°$ (angle in a semicircle). So in right △MNP, $\angle MNP=90°-\angle NMP$, and hence ∠MOP and ∠MNP are both acute.

∠MOP = ∠MNP (angles in the same segment, both subtended by MP); and since MN is a diameter, ∠MON = ∠MPN = 90°, so both these angles are acute.

Let E be a point on side CD produced beyond D. The exterior angle at D is ∠ADE, and ∠ADE and ∠ADC are angles on the straight line CE, so

$$\angle ADE+\angle ADC=180°$$

Since ABCD is cyclic, the opposite angles give

$$\angle ABC+\angle ADC=180°$$

Comparing the two equations, $angleADE=angleABC$: the exterior angle at D equals the interior angle at the opposite vertex B. The same argument works at every vertex.

Both the exterior angle at D and ∠ABC are supplementary to ∠ADC, so they are equal.

Let AB be any chord of a circle with centre O and radius r.

• If AB passes through O, it is a diameter: AB = 2r.
• Otherwise O, A, B form a triangle, and by the triangle inequality $AB<OA+OB=r+r=2r$.

Alternatively, a chord at distance d from the centre has length $2\sqrt{r^2-d^2}\le 2\sqrt{r^2}=2r$, with equality only when d = 0.

For any chord AB, AB ≤ OA + OB = 2r, with equality only when the chord passes through the centre, i.e. when it is a diameter.

Take any chord through A and let d be its distance from O, i.e. the length of the perpendicular OM from O to the chord. Its length is $2\sqrt{r^2-d^2}$, so the chord is shortest when d is largest.

In right △OMA (right angle at M), OA is the hypotenuse, so $d=OM\le OA$. The largest possible value, d = OA, happens exactly when M = A, that is, when the chord is perpendicular to OA at A.

So the chord through A perpendicular to OA is the shortest; its length is $2\sqrt{r^2-O{A}^2}$.

Chord length $=2\sqrt{r^2-d^2}$ is least when d is greatest; since d ≤ OA with equality only for the chord perpendicular to OA at A, that chord is the shortest.

In Fig. 5.30 the base is a diameter with centre O, A is a point on the semicircle, and OA is drawn. The tick marks show that the two halves of the diameter and OA are equal (all radii). The base angles are a and b.

• In △OA(left end): the two sides at O are radii, so it is isosceles and the angle at A in this triangle is also a.
• Similarly, in the right triangle the angle at A is b.

So the angle of the big triangle at A is $a+b$. Adding the three angles of the big triangle:

$$a+b+(a+b)=180°\Rightarrow 2(a+b)=180°\Rightarrow a+b=90°$$

The two isosceles triangles give the angle at A as a + b; the angle sum of the large triangle gives 2(a + b) = 180°, so the angle in the semicircle is a + b = 90°.

A chord perpendicular to a diameter is bisected by it (Theorem 5: the perpendicular from the centre bisects the chord). So AB is the perpendicular bisector of CC' and of DD': C' is the mirror image of C in the line AB, and D' is the mirror image of D.

Take AB along the x-axis with the centre at the origin. Then for some numbers:

$$C=(c_1,c_2),C^{\prime }=(c_1,-c_2),D=(d_1,d_2),D^{\prime }=(d_1,-d_2)$$

$$M=\left(\frac{c_1+d_1}{2},\frac{c_2+d_2}{2}\right),M^{\prime }=\left(\frac{c_1+d_1}{2},-\frac{c_2+d_2}{2}\right)$$

M and M' have the same x-coordinate, so MM' is a vertical segment, i.e. perpendicular to AB (the x-axis). In words: M' is the mirror image of M in AB, and the segment joining a point to its mirror image is perpendicular to the mirror.

Reflection in AB maps C → C′ and D → D′, so it maps the midpoint M of CD to the midpoint M′ of C′D′; hence MM′ ⊥ AB.

In Fig. 5.31 the centre O is joined to the four vertices A, B, C, D. Each of OA, OB, OC, OD is a radius (shown by the tick marks), so the four triangles OAB, OBC, OCD, ODA are isosceles. Let their base angles be w (in △OAB), x (in △OBC), y (in △OCD) and z (in △ODA).

Each angle of the quadrilateral is made of two base angles:

$$\angle A=z+w,\angle B=w+x,\angle C=x+y,\angle D=y+z$$

The angles of a quadrilateral add up to 360°:

$$2(w+x+y+z)=360°\Rightarrow w+x+y+z=180°$$

Therefore

$$\angle A+\angle C=(z+w)+(x+y)=180°,\angle B+\angle D=(w+x)+(y+z)=180°$$

Splitting each angle into base angles of the four isosceles triangles shows ∠A + ∠C = ∠B + ∠D = half of 360° = 180°.