Chapter 5
I’m Up and Down,
and Round and Round
A complete, interactive companion — step-by-step solutions to every question in the chapter.
Think and Reflect — Symmetries of a Circle
A regular polygon with n sides has n rotational symmetries and n lines of reflection symmetry.
Square (n = 4): Rotations by 90°, 180°, 270°, 360°. Lines of reflection: 2 diagonals + 2 perpendicular bisectors of sides.
Regular Pentagon (n = 5): 5 rotational symmetries (multiples of 72°), 5 lines of reflection.
Regular Hexagon (n = 6): 6 rotational symmetries (multiples of 60°), 6 lines of reflection.
The longest chord is the diameter: $\text{Longest} = 2r = 2 \times 5 = 10 \text{ units}$.
For the smallest chord — as a chord shrinks to a single point on the circle, its length approaches 0 but never quite reaches 0. So there is no smallest chord.
The locus is the perpendicular bisector of segment AB.
Proof: If P lies on the perpendicular bisector through midpoint M, then in $\triangle PMA$ and $\triangle PMB$: $AM = BM$, $PM$ common, $\angle PMA = \angle PMB = 90°$. By SAS, $\triangle PMA \cong \triangle PMB$, so $PA = PB$.
Conversely, every point with $PA = PB$ lies on this line.
Think and Reflect — How Many Circles?
The centre of any such circle must be equidistant from A and B, so it must lie on the perpendicular bisector of AB.
Since the perpendicular bisector has infinitely many points, we can draw a circle for each — giving infinitely many circles.
Smallest: Take midpoint of AB as centre — then AB is the diameter. Any smaller circle cannot enclose both A and B.
Largest: As the centre moves farther along the perpendicular bisector, the radius grows without bound — there is no largest circle.
If O lies on the perpendicular bisector at distance $d$ from midpoint M, then by Pythagoras:
As $d$ increases, $OA$ (the radius) also increases.
“More curved” = smaller radius (tighter turn); “less curved” = larger radius (flatter).
Since the radius grows as we move away (Q3), the circle becomes flatter — less curved.
(a) On the boundary: A and B can be anywhere on any side of a square — many orientations and sizes are possible. Hence infinitely many squares.
(b) As corners of the square:
- If AB is a side of the square: 2 squares (one on each side of AB).
- If AB is the diagonal: 1 square (uniquely determined).
Total: $2 + 1 = 3$ squares.
Circumcircles & Three Points
$\angle C = 180° – 70° – 60° = 50°$
All three angles are less than 90° — so △ABC is acute-angled.
For acute triangles, the circumcentre lies inside the triangle.
$\angle A = 100° > 90°$ — so △ABC is obtuse-angled at A.
For obtuse triangles, the circumcentre lies outside the triangle, on the side opposite to the obtuse angle.
O is the circumcentre, so $OA = OB = OC = R$ (the circumradius).
Using Heron’s formula with $s = 10$:
The smallest such circle has AB as its diameter — centre at the midpoint of AB.
Think, Draw and Infer
If $PA = PB$, then P lies on the perpendicular bisector of AB. If $PB = PC$, then P lies on perpendicular bisector of BC.
Both A, B, C lie on the same line $\ell$. So both perpendicular bisectors are perpendicular to $\ell$ — making them parallel (they pass through different midpoints, so they never meet).
Hence no such P exists. So no circle passes through 3 collinear points. A line can intersect a circle in at most 2 points.
Yes! Rotate △ABC about the circumcentre O by any angle. The rotated triangle is congruent to the original and has the same circumcircle (rotation preserves circles through O).
Triangles Formed by Chords
Let AB be any chord of a circle with centre O. Join OA and OB.
Since OA and OB are both radii: $OA = OB = r$.
A triangle with two equal sides is isosceles, so △OAB is isosceles. $\blacksquare$
Let △OAB and △OCD be two such triangles in the same circle, with $AB = CD$.
- $OA = OC = r$ (radii)
- $OB = OD = r$ (radii)
- $AB = CD$ (given)
By the SSS congruence criterion: $\triangle OAB \cong \triangle OCD$. $\blacksquare$
Perpendicular Bisectors of Chords
Let C be the centre, AB the chord, and CM ⊥ AB with M on AB.
In △CMA and △CMB:
- $CA = CB = r$ (radii — hypotenuses)
- $CM = CM$ (common side)
- $\angle CMA = \angle CMB = 90°$
By RHS congruence: $\triangle CMA \cong \triangle CMB$, hence $AM = BM$. $\blacksquare$
In an isosceles triangle, the altitude from the apex (A) is also the perpendicular bisector of the base BC.
By Theorem 5, the perpendicular bisector of any chord passes through the centre O of the circle.
Since BC is a chord and AD (the altitude) is its perpendicular bisector, AD passes through O. $\blacksquare$
For chord 6 cm (half = 3): $d_1 = \sqrt{5^2 – 3^2} = \sqrt{16} = 4$ cm
For chord 8 cm (half = 4): $d_2 = \sqrt{5^2 – 4^2} = \sqrt{9} = 3$ cm
Since chords are on opposite sides of the centre, total distance:
Equal Chords & Pythagoras
Let AB and FG be equal chords. Let E, H be midpoints of AB, FG respectively, with C the centre.
Since AB = FG and E, H are midpoints: $AE = FH$.
$CA = CF = r$ (radii). Apply Pythagoras in right triangles CEA and CHF:
Since $AE = FH$: $CE^2 = CH^2 \Rightarrow CE = CH$. $\blacksquare$
In right triangles CEA and CHF:
- $\angle CEA = \angle CHF = 90°$
- $CA = CF = r$ (hypotenuses)
- $CE = CH$ (given)
By RHS: $\triangle CEA \cong \triangle CHF$ ⇒ $AE = FH$.
By Theorem 5, E and H are midpoints, so $AB = 2AE = 2FH = GF$. $\blacksquare$
Apply Pythagoras in △CEA and △CHF:
Given $CE = CH$: $AE^2 = FH^2$ ⇒ $AE = FH$.
So $AB = 2AE = 2FH = GF$. $\blacksquare$
Distance of Chords from the Centre
Using $\text{Chord} = 2\sqrt{r^2 – d^2}$:
Let AB be the chord, O the centre, M the foot of perpendicular from O to AB. By Theorem 5, M is the midpoint of AB.
△OMA is right-angled at M with $OA = r$ (hypotenuse), $OM = d$.
By Pythagoras:
Since M is the midpoint: $AB = 2 \cdot AM = 2\sqrt{r^2 – d^2}$. $\blacksquare$
No. Let distance of CD be $d$, so distance of AB is $2d$. Then:
For $CD = 2 \cdot AB$ we would need $r^2 – d^2 = 4(r^2 – 4d^2)$, giving $r^2 = 5d^2$ — only a special case, not general.
Counter-example: $r = 5, d = 1$:
$AB = 2\sqrt{21} \approx 9.17$, $CD = 2\sqrt{24} \approx 9.80$ — clearly $CD \neq 2 AB$.
Angles Subtended by Arcs
△AOB has $OA = OB = 12$ cm and $\angle AOB = 60°$.
Base angles: $\angle OAB = \angle OBA = \dfrac{180° – 60°}{2} = 60°$.
All three angles are 60° → △AOB is equilateral.
(i) No. By Theorem 9 (corollary), all points on the same arc subtend equal angles at chord AB. So $\angle AXB = \angle AYB$ always.
(ii) Not necessarily. If AB is a diameter, then any point on either arc gives 90°. So X, Y can lie on opposite arcs and still both equal 90°.
(iii) Yes. By Theorem 10, if AB subtends equal angles at two points on the same side, the four points are concyclic. So Y lies on the circle through A, B, X.
In a cyclic quadrilateral, opposite angles sum to 180° (Theorem 11):
Mixed Exercises
By Theorem 9: the angle at the centre is double the angle at any point on the circle (outside the arc).
Radius $r = 13$ cm. Half-chord = 12 cm.
Let AB be a chord, O the centre. Then $OA = OB = r$ (both radii).
Since O is equidistant from A and B, by the locus property of perpendicular bisectors, O lies on the perpendicular bisector of AB.
Hence the perpendicular bisector of AB passes through O. $\blacksquare$
Centre O is midpoint of AB, so $OA = OB = OC = r$.
△OAC is isosceles: $\angle OAC = \angle OCA = \alpha$.
△OBC is isosceles: $\angle OBC = \angle OCB = \beta$.
In △ABC: $\alpha + \beta + \angle ACB = 180°$ and $\angle ACB = \alpha + \beta$.
So $2\angle ACB = 180°$ ⇒ $\angle ACB = 90°$.
Opposite angles in cyclic quadrilateral sum to 180°:
∠P and ∠R are opposite angles:
$\angle P = 2(38) + 10 = 86°$, $\angle R = 3(38) – 20 = 94°$. Check: $86° + 94° = 180°$ ✓
Half-chord = 8. By Pythagoras:
Use Brahmagupta’s formula with semi-perimeter $s = \dfrac{5+5+12+12}{2} = 17$:
Since opposite angles of a cyclic quadrilateral sum to 180°, either all four are 90° (rectangle) or two are obtuse and two are acute.
- Rectangle/square: The diagonals are diameters, so the centre is at the intersection of the diagonals — clearly inside.
- One diagonal subtends 90°: That diagonal is a diameter, so the centre lies on it (at its midpoint). The centre is inside if that midpoint lies inside the quadrilateral.
- General case: Find the centre as the intersection of the perpendicular bisectors of any two sides — then check by inspection whether it lies inside the quadrilateral.
Quick rule: The centre lies inside iff every side (as a chord) has the rest of the quadrilateral on the side of its major arc.
Let chords AB and CD intersect at P, with $AB = CD$. Drop perpendiculars OM, ON from centre O to AB, CD.
By Theorem 5: M, N are midpoints. By Theorem 6: $OM = ON$ (equal chords are equidistant).
In right triangles OMP and ONP:
- $OM = ON$
- $OP$ common
- $\angle OMP = \angle ONP = 90°$
By RHS: $\triangle OMP \cong \triangle ONP \Rightarrow MP = NP$.
Since $AM = BM = CN = DN = AB/2$, the segments $AP, BP$ are $AM \pm MP$ and $CP, DP$ are $CN \pm NP$. With $AM = CN$ and $MP = NP$, the pairs match. $\blacksquare$
First find the radius:
Construction:
- Draw chord AB of length 6 cm.
- Construct perpendicular bisector of AB; let it meet AB at M.
- Mark point O on this bisector at distance 3 cm from M.
- With O as centre, OA as radius, draw the circle.
Let parallelogram ABCD be cyclic (inscribed in a circle).
Cyclic ⇒ $\angle A + \angle C = 180°$.
Parallelogram ⇒ $\angle A = \angle C$ (opposite angles equal).
So $2\angle A = 180°$ ⇒ $\angle A = 90°$. Similarly all four angles are 90°.
Hence ABCD is a rectangle. $\blacksquare$
Since $\angle B = 90°$ and $\angle B$ is inscribed in the circle subtending chord AC: by the converse of “angle in semicircle = 90°”, AC must be a diameter.
Similarly, BD is also a diameter (since $\angle A = 90°$ subtends BD).
Two diameters of a circle both pass through the centre, so their intersection is the centre. $\blacksquare$
For a chord of length $2\ell$ in a circle of radius $r$, the midpoint M satisfies $OM = \sqrt{r^2 – \ell^2}$ — a fixed distance.
So all midpoints lie at the same distance from O — they form a concentric circle of radius $\sqrt{r^2 – \ell^2}$.
Since $AB = AC$ (equal chords), by Theorem 6, they are equidistant from O.
So the perpendicular distance from O to line AB equals the perpendicular distance from O to line AC.
A point equidistant from the two sides of an angle lies on its bisector.
Hence O lies on the bisector of $\angle BAC$. $\blacksquare$
Let radius = $r$. Distances:
Same side, gap = 7: $d_1 – d_2 = 7$.
Let $a = d_1, b = d_2$. Then $a – b = 7$, and $a^2 – b^2 = 119$.
$(a-b)(a+b) = 119 \Rightarrow a + b = 17$. Solving: $a = 12, b = 5$.
Join O to two adjacent vertices A, B. △OAB has $OA = OB = r$ and $\angle AOB = 360°/6 = 60°$.
Base angles: $(180°-60°)/2 = 60°$. All angles are 60° ⇒ △OAB is equilateral.
So side $AB = r$. Distance from O to AB = height of equilateral triangle:
In cyclic quadrilateral MNOP, both N and O lie on the same arc with respect to chord MP. By the property “angles in the same segment are equal”:
Additionally, since MN is a diameter, $\angle MPN = 90°$ (angle in a semicircle).
Extend CD beyond D to a point E. The exterior angle at D is $\angle ADE$.
Linear pair: $\angle ADC + \angle ADE = 180°$ … (i)
Cyclic property: $\angle ADC + \angle ABC = 180°$ … (ii)
Subtracting: $\angle ADE = \angle ABC$. $\blacksquare$
Let AB be any chord, O the centre with radius $r$.
By the triangle inequality on points O, A, B:
Equality holds iff O lies on segment AB — i.e., AB is a diameter. $\blacksquare$
Any chord through A has midpoint M with $OM \perp$ chord (Theorem 4). Chord length:
To minimise chord, maximise $OM$.
For any chord through A: $OM \le OA$ (perpendicular is the shortest distance).
Equality holds when $M = A$ ⇒ $OA \perp$ chord. So the shortest chord is perpendicular to OA. $\blacksquare$
In the figure, O is the midpoint of the diameter, and A is on the semicircle. Join OA. Since OA, OB, OC are all radii:
- △OAB is isosceles ⇒ base angles at A (left part) and at B both equal $a$.
- △OAC is isosceles ⇒ base angles at A (right part) and at C both equal $b$.
Sum of triangle angles: $a + b + \angle A = 180°$. But $\angle A = a + b$.
So $\angle A = 90°$. $\blacksquare$
Since CC′ ⊥ diameter AB, by Theorem 5, AB bisects CC′ — so C and C′ are mirror images across AB.
Similarly, D and D′ are mirror images across AB.
Therefore segment CD is the reflection of segment C′D′ across AB. The midpoint M of CD reflects to the midpoint M′ of C′D′.
The line joining a point and its mirror image (across a line) is always perpendicular to that line.
Hence MM′ ⊥ AB. $\blacksquare$
O is joined to all four vertices, creating four isosceles triangles (two sides = radius).
Label base angles: $\alpha$ in △OAB, $\beta$ in △OBC, $\gamma$ in △OCD, $\delta$ in △ODA.
Interior angles of quadrilateral:
- $\angle A = \alpha + \delta$
- $\angle B = \alpha + \beta$
- $\angle C = \beta + \gamma$
- $\angle D = \gamma + \delta$
Sum of all four = $2(\alpha + \beta + \gamma + \delta) = 360°$.
So $\angle A + \angle C = (\alpha+\delta) + (\beta+\gamma) = 180°$. Similarly $\angle B + \angle D = 180°$. $\blacksquare$