Answer: B
A circle is the set (locus) of all points in a plane at a fixed distance from a fixed point called the centre.

Answer: C
The longest chord passes through the centre, so it is a diameter of length 2r.

Answer: B
The chapter notes that every diameter is a line of reflection symmetry of a circle.

Answer: D
Their centres may be any points on the perpendicular bisector of AB, giving infinitely many circles.

Answer: B
The smallest circle occurs when AB is a diameter, so radius=d/2.

Answer: B
The perpendicular bisectors of two sides meet at one point, the unique circumcentre.

Answer: A
For an acute triangle, the circumcentre is inside the triangle.

Answer: B
The chapter states that the circumcentre of a right triangle is the midpoint of its hypotenuse.

Answer: B
Equal chords subtend equal central angles.

Answer: C
This is the converse theorem: equal central angles subtend equal chords.

Answer: B
The centre-to-midpoint line is perpendicular to the chord.

Answer: A
The perpendicular from the centre bisects the chord.

Answer: A
Equal chords are equidistant from the centre.

Answer: B
The chapter proves that a longer chord lies closer to the centre.

Answer: B
The perpendicular bisects the chord; half-chord=√(r²−d²), so chord=2√(r²−d²).

Answer: B
Chord=2√(49−36)=2√13 cm.

Answer: B
The two radii and chord form an equilateral triangle when the central angle is 60°, so the chord is 12 cm.

Answer: B
The central angle is twice the angle at the circle, so the angle is 50°.

Answer: C
A diameter subtends 180° at the centre and therefore 90° at the circumference.

Answer: A
They are each half the same central angle.

Answer: C
Opposite angles of a cyclic quadrilateral add to 180°.

Answer: B
The converse theorem states that the four vertices are concyclic.

Answer: C
(2x+10)+(3x−20)=180 ⇒5x−10=180 ⇒x=38.

Answer: A
Triangles AOB and AOC are congruent by SSS, so ∠BAO=∠OAC.

Answer: B
Each central angle is 60°, so each chord side equals the radius.

Answer: C
Half a side is r/2. By Pythagoras, distance=√(r²−r²/4)=√3r/2.

Answer: C
Distances from centre are √(25−3²)=4 and √(25−4²)=3. Opposite sides ⇒ total=7 cm.

Answer: B
Let distances be √(r²−25) and √(r²−144); their difference is 7. Solving gives r=13 cm.

Answer: C
Fixed chord length gives fixed perpendicular distance from the centre, so all midpoints lie on a circle centred at the same centre.

Answer: B
Opposite angles of a parallelogram are equal, while in a cyclic quadrilateral they are supplementary. Hence each is 90°, so it is a rectangle.

A chord is a line segment joining two points on the circle.

A chord passing through the centre of the circle.

Infinitely many; every diameter is a line of reflection symmetry.

The perpendicular bisector of AB.

Exactly one.

Outside the triangle.

Equal chords of a circle subtend equal angles at the centre.

The perpendicular distance from the centre to the chord.

A diameter.

2√(25−9)=8 cm.

42°.

90°.

112°.

They are supplementary, so the quadrilateral is cyclic.

Half-chord=6; distance=√(100−36)=8 cm.

Half-chord=√(13²−5²)=√144=12. Hence chord=24 cm.

The smallest circle occurs when AB is a diameter. Radius=AB/2=5 cm.

No. The centre can move arbitrarily far along the perpendicular bisector of AB, making the radius arbitrarily large.

(i) All acute ⇒ circumcentre inside. (ii) Obtuse triangle ⇒ circumcentre outside.

Equal chords subtend equal angles at the centre, so ∠COD=74°.

Central angle=2×56°=112°.

∠C=180−72=108°. ∠D=180−109=71°.

Half-chord=8. r²=8²+6²=100, so r=10 cm.

Side=8 cm. Distance=√(8²−4²)=√48=4√3 cm.

The centre would have to lie on perpendicular bisectors of AB and BC. For collinear A,B,C these perpendicular bisectors are parallel distinct lines, so they do not meet.

∠ACB=90°. Hence ∠ABC=180−90−34=56°.

The exterior angle of a cyclic quadrilateral equals the interior opposite angle, so ∠B=68°.

Half-chord=5. Distance=√(169−25)=12 cm.

The 12 cm chord lies closer, because the longer chord of a circle is nearer the centre.

No. Chord length is 2√(r²−d²), which depends non-linearly on d; changing d by a factor does not give the same factor change in chord length.

Let AB=DE in a circle with centre O. OA=OB=OD=OE (radii). Thus ΔOAB≅ΔODE by SSS. Hence ∠AOB=∠DOE.

If ∠AOB=∠DOE and OA=OD, OB=OE (radii), then ΔOAB≅ΔODE by SAS. Hence AB=DE.

OA=OB, AM=MB and OM is common. Thus ΔOMA≅ΔOMB by SSS. Therefore ∠OMA=∠OMB; they form a linear pair, so each is 90°. Hence OM⊥AB.

Let OM⊥AB. In right triangles OMA and OMB, OA=OB (radii), OM is common and both have right angles. By RHS, ΔOMA≅ΔOMB, so AM=MB.

Let equal chords AB,CD have perpendicular distances OM,ON. The perpendiculars bisect the chords, so AM=CN. In right triangles OMA,ONC, OA=OC=r and AM=CN. Thus OM²=r²−AM²=r²−CN²=ON², hence OM=ON.

Let OM=ON be perpendicular distances to chords AB,CD. In right triangles OMA,ONC, OA=OC=r and OM=ON. Hence AM²=r²−OM²=r²−ON²=CN², so AM=CN. Therefore AB=2AM=2CN=CD.

Let half-chords be a>b and distances d₁,d₂. Since r²=d₁²+a²=d₂²+b² and a²>b², we must have d₁²<d₂², hence d₁<d₂.

For 6 cm chord: half=3, distance=√(25−9)=4. For 8 cm chord: half=4, distance=√(25−16)=3. Opposite sides ⇒ midpoint distance=4+3=7 cm.

Let distances be d₁ for 10 cm chord and d₂ for 24 cm chord, d₁−d₂=7. Then r²=d₁²+25=d₂²+144. Thus d₁²−d₂²=119 ⇒(d₁−d₂)(d₁+d₂)=119 ⇒7(d₁+d₂)=119 ⇒d₁+d₂=17. Hence d₁=12,d₂=5 and r²=144+25=169, so r=13 cm.

The centre must be equidistant from A,B, so it lies on the perpendicular bisector of AB; similarly it lies on that of AC. Since the points are non-collinear, these bisectors intersect at one unique point O. The circle centred at O with radius OA passes through A,B,C.

The arc AB not containing C or D is a semicircle and subtends 180° at the centre. By the central-angle theorem, each angle at C and D is half of 180°, i.e. 90°.

In cyclic ABCD, ∠ABC+∠ADC=180°. The exterior angle at D formed by extending CD is 180°−∠ADC. Therefore exterior angle=∠ABC.

Central angle for each side=360°/6=60°. The triangle formed by two radii and one side is equilateral, so side=r. Perpendicular from centre bisects side, giving distance=√(r²−(r/2)²)=√3r/2.

For chord 12: d=√(100−36)=8. For chord 16: d=√(100−64)=6. If on same side, separation=2 cm; if on opposite sides, separation=14 cm.

Opposite angles are supplementary. Let ∠A=2k, ∠C=3k. Then 5k=180°, k=36°. Thus ∠A=72°, ∠C=108°.

Let A,B,C be non-collinear. A centre O of any circle through A,B must lie on the perpendicular bisector of AB; similarly, because OA=OC, it must lie on the perpendicular bisector of AC. These two bisectors intersect at exactly one point because AB and AC are not the same line and are not parallel in the required construction. Call the intersection O. Then OA=OB and OA=OC, so OA=OB=OC. Therefore the circle with centre O and radius OA passes through all three points. Since the two bisectors have only one intersection, no second centre and hence no second circle is possible.

Every centre must lie on the perpendicular bisector of AB, and every point on that bisector is equidistant from A,B, so infinitely many centres/circles exist. If M is midpoint and centre O is on the bisector, OA²=OM²+(d/2)². This is least when OM=0, giving radius d/2. As OM can grow without bound, OA can grow without bound; hence there is no largest radius.

Let OM=d be perpendicular to chord AB. Then M is its midpoint, so AM=AB/2. In right triangle OMA, r²=d²+AM². Thus AM=√(r²−d²) and AB=2√(r²−d²). For r=17,d=8: AB=2√(289−64)=2√225=30.

Let arc AB subtend ∠AOB at centre O and ∠ADB at point D outside the arc. Join OD and extend it to E. Since OA=OD and OB=OD, triangles AOD and BOD are isosceles. Using the exterior-angle theorem, the relevant central component angles are twice the corresponding angles at D. Adding (or subtracting, depending on the position of E) the component relations gives ∠AOB=2∠ADB.

Draw the unique circle through non-collinear A,B,C. Suppose D is not on it. If D is outside, let AD meet the circle at E. Then C,E are in the same segment, so ∠ACB=∠AEB. But ∠AEB is an exterior angle of triangle BED, so ∠AEB>∠ADB, contradicting ∠ACB=∠ADB. A similar contradiction occurs if D is inside. Hence D lies on the circle and all four points are concyclic.

In cyclic ABCD, ∠BAD is half the central angle subtended by arc BCD, while ∠BCD is half the central angle subtended by the other arc BAD. These two central angles together make 360°. Therefore ∠BAD+∠BCD=1/2×360°=180°. Similarly the other opposite pair is supplementary.

Take ABCD with ∠A+∠C=180°. Draw the circle through A,B,D. If C were not on it, let line CD meet the circle at E. In cyclic ABED, ∠A+∠BED=180°, while given ∠A+∠BCD=180°, so ∠BED=∠BCD. But if C lies outside or inside the circle, one of these is an exterior angle of an appropriate triangle and must be larger than the other, a contradiction. Thus C lies on the circle and ABCD is cyclic.

Let ABCD be cyclic with AB=BC=5 and CD=DA=12. Diagonal AC divides it into isosceles triangles ABC and ADC. Let ∠ABC=2u. Then ∠ADC=180°−2u. In triangle ABC, AC=10 sin u; in triangle ADC, AC=24 cos u. Hence 10 sin u=24 cos u ⇒tan u=12/5, so sin u=12/13, cos u=5/13. Therefore AC=120/13. Heights to AC are √(25−(60/13)²)=25/13 and √(144−(60/13)²)=144/13. Total height=13. Area=1/2×(120/13)×13=60 square units.

For any chord through A, its length decreases as its perpendicular distance from O increases. Among all lines through A, the greatest possible perpendicular distance from O is OA, achieved exactly by the line through A perpendicular to OA. Hence that chord is shortest. Its distance from O is 6, so length=2√(10²−6²)=2√64=16.

For any such chord, the perpendicular from centre O passes through midpoint M. In right triangle OMA, OA=R and half-chord AM=a. Hence OM=√(R²−a²), a constant. Therefore every midpoint lies on the circle centred at O with radius √(R²−a²). Conversely, any point M at this distance determines a perpendicular chord of half-length a, so the entire locus is that concentric circle.