Circles

(i) The centre of a circle lies in interior of the circle. (exterior/interior)

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle. (exterior/interior)

(iii) The longest chord of a circle is a diameter of the circle.

(iv) An arc is a semicircle when its ends are the ends of a diameter.

(v) Segment of a circle is the region between an arc and the chord of the circle.

(vi) A circle divides the plane, on which it lies, in three parts.

(i) Line segment joining the centre to any point on the circle is a radius of the circle.
True. Because all points on the circle are at equal distance from the centre. This constant distance is called radius.

(ii) A circle has only finite number of equal chords.
False. A circle has infinite points, so we can draw infinitely many equal chords.

(iii) If a circle is divided into three equal arcs, each is a major arc.
False. For an arc to be major, it must be larger than a semicircle. Three equal arcs will each read 120°, which is less than 180°.

(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
True. Because diameter = 2 × radius.

(v) Sector is the region between the chord and its corresponding arc.
False. That region is called a segment. Sector is the region between two radii and an arc.

(vi) A circle is a plane figure.
True. Since circle is a collection of points in a two-dimensional plane.

Let there be two congruent circles with centres O and O'.
Assume AB is a chord of the first circle and CD is a chord of the second such that AB = CD.
In ΔAOB and ΔC'O'D':
OA = O'C' (Radii of congruent circles)
OB = O'D' (Radii of congruent circles)
AB = CD (Given)
Therefore, ΔAOB ≅ ΔC'O'D' (By SSS Congruence Rule)
Hence, ∠AOB = ∠C'O'D' (By CPCT)
This proves that equal chords of congruent circles subtend equal angles at their centres.

Let there be two congruent circles with centres O and O'.
Assume AB and CD are chords such that ∠AOB = ∠C'O'D'.
In ΔAOB and ΔC'O'D':
OA = O'C' (Radii of congruent circles)
OB = O'D' (Radii of congruent circles)
∠AOB = ∠C'O'D' (Given)
Therefore, ΔAOB ≅ ΔC'O'D' (By SAS Congruence Rule)
Hence, AB = CD (By CPCT)
This proves that if chords subtend equal angles at the centres, the chords are equal.

When you draw different pairs of circles, they can intersect in:
- 0 points (Non-intersecting circles)
- 1 point (Circles touching each other externally or internally)
- 2 points (Intersecting circles)
Hence, the maximum number of common points is 2.

Steps to find the centre:
1. Take any three points A, B, and C on the given circle.
2. Join AB and BC to form two chords.
3. Draw the perpendicular bisector of chord AB.
4. Draw the perpendicular bisector of chord BC.
5. Let the two perpendicular bisectors intersect at a point O. Point O is the centre of the circle.
(Reason: Perpendicular bisectors of chords pass through the centre).

Let two circles with centres O and O' intersect at points A and B.
AB is the common chord. Join OA, OB, O'A, and O'B. Join OO'.
In ΔAOO' and ΔBOO':
OA = OB (Radii of large circle)
O'A = O'B (Radii of small circle)
OO' = OO' (Common side)
Therefore, ΔAOO' ≅ ΔBOO' (By SSS congruence)
So, ∠AOO' = ∠BOO' (By CPCT)

Let OO' intersect AB at M.
In ΔAOM and ΔBOM:
OA = OB (Radii)
∠AOM = ∠BOM (Proved above)
OM = OM (Common)
Therefore, ΔAOM ≅ ΔBOM (By SAS congruence)
So, AM = MB and ∠OMA = ∠OMB = 90° (linear pair is 180°).
Hence, OO' is the perpendicular bisector of AB.

Let O be the centre of circle with radius 5 cm, and P be the centre of circle with radius 3 cm. Distance OP = 4 cm.
Let the circles intersect at A and B. AB is the common chord intersecting OP at M.
Notice that 3² + 4² = 9 + 16 = 25 = 5².
This means ΔOPA is a right-angled triangle with ∠OPA = 90°.
Since OP is perpendicular to the common chord from the centre of the smaller circle, the smaller circle's centre P lies on the common chord AB.
Thus, AB is a diameter of the smaller circle.
Length of common chord AB = 2 × radius of smaller circle = 2 × 3 = 6 cm.

Let AB and CD be two equal chords (AB = CD) of a circle with centre O, intersecting at point E.
Draw OM ⊥ AB and ON ⊥ CD. Join OE.
In ΔOME and ΔONE:
OM = ON (Equal chords are equidistant from the centre)
∠OME = ∠ONE = 90°
OE = OE (Common)
Therefore, ΔOME ≅ ΔONE (RHS Congruence)
Thus, ME = NE (CPCT) --- (i)

We know AB = CD. So, half of equal chords are equal:
AM = CN --- (ii)
Adding (i) and (ii): AM + ME = CN + NE ⇒ AE = CE.
Also, AB - AE = CD - CE ⇒ BE = DE.
Hence proved.

Let AB and CD be equal chords intersecting at E inside the circle with centre O. Join OE.
Draw OM ⊥ AB and ON ⊥ CD.
From the previous proof, ΔOME ≅ ΔONE (By RHS).
Therefore, ∠MEO = ∠NEO (By CPCT).
This means OE makes equal angles with the chords.

Let O be the common centre. The line intersects the outer circle at A, D and the inner circle at B, C.
Draw OM ⊥ AD.
Since OM is perpendicular to chord BC of the inner circle, it bisects it:
BM = CM --- (i)
Since OM is perpendicular to chord AD of the outer circle, it bisects it:
AM = DM --- (ii)
Subtracting (i) from (ii):
AM - BM = DM - CM ⇒ AB = CD.
Hence proved.

Let R, S, M represent Reshma, Salma, Mandip. OS is the radius = 5 m. RS = SM = 6 m.
Draw OM ⊥ RS. In ΔORS, Area = ½ × RS × OM OR ½ × OS × RN (where RN ⊥ OS).
Since OM bisects RS, RM = 3 m.
In ΔORM, OM = √(OR² - RM²) = √(5² - 3²) = 4 m.
Area of ΔORS = ½ × RS × OM = ½ × 6 × 4 = 12 m².
Also, Area of ΔORS = ½ × OS × RN.
12 = ½ × 5 × RN ⇒ RN = 24 / 5 = 4.8 m.
Since OS acts as perpendicular bisector of chord RM, total distance RM = 2 × RN = 2 × 4.8 = 9.6 m.
Distance between Reshma and Mandip is 9.6 m.

Let A, S, D represent Ankur, Syed, and David. They sit at equal distances, so ΔASD is an equilateral triangle inscribed in the circle of radius R = 20 m.
Let the side of the triangle be a.
In an equilateral triangle, Circumradius R = a / √3.
20 = a / √3.
a = 20√3 m.
Length of the string = 20√3 m (approx 34.64 m).

Total central angle subtended by arc AC is ∠AOC.
∠AOC = ∠AOB + ∠BOC = 60° + 30° = 90°.
The angle subtended by an arc at the centre is double the angle subtended by it at any remaining part of the circle.
∠ADC = ½ ∠AOC = ½ × 90° = 45°.
Thus, ∠ADC = 45°.

Let chord AB = radius r. Join OA and OB.
Since OA = OB = AB = r, ΔAOB is an equilateral triangle.
Hence, ∠AOB = 60°.
Angle subtended by arc AB on the major arc (say point M) = ½ ∠AOB = ½ × 60° = 30°.
Let point N be on the minor arc. AMBN forms a cyclic quadrilateral.
In a cyclic quadrilateral, opposite angles sum to 180°.
∠ANB + ∠AMB = 180°
∠ANB + 30° = 180°
∠ANB = 150°.
Angle on major arc = 30°, Angle on minor arc = 150°.

In cyclic quadrilateral, let S be a point on the major arc PR.
Then ∠PSR + ∠PQR = 180° ⇒ ∠PSR = 180° - 100° = 80°.
Angle subtended by arc PQR at centre (reflex ∠POR) is 2 × ∠PQR = 200°.
Inner ∠POR = 360° - 200° = 160° (or 2 × ∠PSR = 160°).
In ΔOPR, OP = OR (radii), so ∠OPR = ∠ORP.
∠OPR + ∠ORP + ∠POR = 180°
2∠OPR + 160° = 180°
2∠OPR = 20° ⇒ ∠OPR = 10°.

In ΔABC, sum of angles is 180°.
∠BAC = 180° - (∠ABC + ∠ACB) = 180° - (69° + 31°) = 180° - 100° = 80°.
Angles in the same segment of a circle are equal.
Therefore, ∠BDC = ∠BAC = 80°.

In ΔCDE, exterior angle ∠BEC = ∠CDE + ∠ECD.
130° = ∠CDE + 20°.
∠CDE = 110°.
Now, ∠BDC (which is same as ∠CDE) = 110°.
Angles in the same segment are equal, so ∠BAC = ∠BDC = 110°.

Angles in the same segment: ∠DAC = ∠DBC = 70°.
Therefore, ∠DAB = ∠DAC + ∠CAB = 70° + 30° = 100°.
In cyclic quad ABCD, opposite angles add to 180°.
∠BCD + ∠DAB = 180° ⇒ ∠BCD = 180° - 100° = 80°.
In ΔABC, if AB = BC, then ∠BCA = ∠BAC = 30°.
∠ECD = ∠BCD - ∠BCA = 80° - 30° = 50°.

Let ABCD be the cyclic quadrilateral, with diagonals AC and BD being diameters.
Since AC is a diameter, the angle in a semicircle is a right angle. So, ∠ABC = 90° and ∠ADC = 90°.
Since BD is a diameter, ∠BAD = 90° and ∠BCD = 90°.
Since all four interior angles are 90°, the cyclic quadrilateral ABCD is a rectangle.

Let ABCD be a trapezium where AB || CD and non-parallel sides AD = BC.
Draw perpendiculars AM and BN on DC.
In ΔAMD and ΔBNC:
AD = BC (Given)
∠AMD = ∠BNC = 90°
AM = BN (Distance between parallel lines is same)
Therefore, ΔAMD ≅ ΔBNC (RHS Congruence).
So, ∠D = ∠C (CPCT).
Since AB || CD, interior consecutive angles sum to 180°: ∠A + ∠D = 180°.
Substitute ∠D with ∠C: ∠A + ∠C = 180°.
Since the opposite angles sum to 180°, ABCD is a cyclic trapezium.

In circle 1 (through A, B, C, P), angles in the same segment are equal:
∠ACP = ∠ABP --- (i)
In circle 2 (through B, C, D, Q), angles in the same segment are equal:
∠QCD = ∠QBD --- (ii)
Since ABD and PBQ are straight intersecting lines at B, vertically opposite angles are equal:
∠ABP = ∠QBD --- (iii)
From (i), (ii), and (iii), we get ∠ACP = ∠QCD. Hence Proved.

Let ABC be a triangle. Let circles be drawn with AB and AC as diameters.
Let the two circles intersect at A and D. Join AD.
Since AB is a diameter, the angle in a semicircle is a right angle: ∠ADB = 90°.
Since AC is a diameter, the angle in a semicircle is a right angle: ∠ADC = 90°.
Adding both angles: ∠ADB + ∠ADC = 90° + 90° = 180°.
This means B, D, and C lie on a straight line. Thus, D lies on the third side BC.

Since ΔABC and ΔADC are right triangles with common hypotenuse AC,
∠ABC = 90° and ∠ADC = 90°.
Both triangles can be inscribed in a circle with diameter AC.
Therefore, points A, B, C, D are concyclic (lie on the same circle).
Now, chord CD subtends angles at points A and B on the remaining circle.
Angles in the same segment are equal. Hence, ∠CAD = ∠CBD.

Let ABCD be a cyclic parallelogram.
In a cyclic quadrilateral, opposite angles add to 180°. So, ∠A + ∠C = 180°.
In a parallelogram, opposite angles are equal. So, ∠A = ∠C.
Therefore, ∠A + ∠A = 180° ⇒ 2∠A = 180° ⇒ ∠A = 90°.
Similarly, ∠B = ∠D = 90°.
A parallelogram with one angle 90° is a rectangle. Hence proved.

Let AB be the chord, O be the centre. Draw OM ⊥ AB.
Perpendicular from centre bisects the chord. So, AM = MB = 5 cm.
In right ΔOMA, OM² + AM² = OA²
OM² + 5² = 13² ⇒ OM² + 25 = 169 ⇒ OM² = 144.
OM = 12 cm.
Distance of the chord from the centre is 12 cm.

Let the chord be AB and centre be O. Draw ON ⊥ AB. ON = 8 cm.
In right ΔONA, OA² = ON² + AN²
17² = 8² + AN² ⇒ 289 = 64 + AN² ⇒ AN² = 225.
AN = 15 cm.
Length of chord AB = 2 × AN = 2 × 15 = 30 cm.

Radius = 13 / 2 = 6.5 cm.
Let AB = 12 cm, CD = 5 cm. Draw OM ⊥ AB, ON ⊥ CD.
AM = 6 cm. In ΔOMA, OM = √(6.5² - 6²) = √(42.25 - 36) = √6.25 = 2.5 cm.
CN = 2.5 cm. In ΔONC, ON = √(6.5² - 2.5²) = √(42.25 - 6.25) = √36.00 = 6 cm.
Since chords are on opposite sides, Total Distance = OM + ON = 2.5 + 6 = 8.5 cm.

Let chords be AB = 8 cm, CD = 6 cm. Draw OP ⊥ AB and OQ ⊥ CD. OQ - OP = 1 cm.
AP = 4 cm, CQ = 3 cm. Let OP = x, then OQ = x + 1.
Both OA and OC are radius r.
r² = x² + 4² = x² + 16 (from ΔOPA)
r² = (x+1)² + 3² = x² + 2x + 1 + 9 = x² + 2x + 10 (from ΔOQC)
x² + 16 = x² + 2x + 10 ⇒ 2x = 6 ⇒ x = 3 cm.
r² = 3² + 16 = 9 + 16 = 25 ⇒ r = 5 cm.
The radius of the circle is 5 cm.

Opposite angles of a cyclic quadrilateral sum up to 180°.
∠A + ∠C = 180° ⇒ (x+y+10) + (x+y-30) = 180 ⇒ 2(x+y) - 20 = 180 ⇒ 2(x+y) = 200 ⇒ x+y = 100 --- (i)
∠B + ∠D = 180° ⇒ (y+20) + (x+y) = 180 ⇒ x + 2y = 160 --- (ii)
Substitute x = 100 - y into (ii):
(100 - y) + 2y = 160 ⇒ 100 + y = 160 ⇒ y = 60.
Put y = 60 in (i): x + 60 = 100 ⇒ x = 40.
Hence, x = 40 and y = 60.

Since chord AB = radius, OA = OB = AB. So ΔAOB is an equilateral triangle.
Central angle ∠AOB = 60°.
The angle subtended by the arc at any point on the major segment is half the angle subtended at the centre.
∠ACB = ½ ∠AOB = ½ × 60° = 30°.
The angle is 30°.

Equal chords subtend equal angles at the centre.
Since PQ = RS, ∠ROS = ∠POQ.
Therefore, ∠ROS = 70°.

Half of chord = 24 / 2 = 12 cm. Distance = 5 cm.
Radius r = √(12² + 5²) = √(144 + 25) = √169 = 13 cm.
Diameter = 2 × 13 = 26 cm.

In an equilateral triangle, each interior angle is 60°.
So ∠BAC = 60°.
The angle subtended by arc BC at the centre is double the angle subtended by it at any remaining part of the circle.
∠BOC = 2 × ∠BAC = 2 × 60° = 120°.
Therefore, ∠BOC = 120°.

Since ABCD is a cyclic quad, ∠ADC + ∠ABC = 180°.
140° + ∠ABC = 180° ⇒ ∠ABC = 40°.
Since AB is a diameter, angle in the semicircle is 90°. Hence ∠ACB = 90°.
In ΔABC: ∠BAC + ∠ABC + ∠ACB = 180°
∠BAC + 40° + 90° = 180°
∠BAC = 180° - 130° = 50°.
Therefore, ∠BAC = 50°.

Since AD is a diameter of the first circle, ∠ABD = 90° (angle in a semicircle).
Since AC is a diameter of the second circle, ∠ABC = 90°.
Therefore, ∠DBC = ∠ABD + ∠ABC = 90° + 90° = 180°.
Since ∠DBC is 180°, D, B, and C lie on a straight line.
Hence, B lies on the line segment DC.

In ΔOAB, OA = OB (radii). Thus ∠OBA = ∠OAB = 35°.
In ΔOBC, OB = OC (radii). Thus ∠OBC = ∠OCB = 40°.
Total ∠ABC = ∠OBA + ∠OBC = 35° + 40° = 75°.
Angle at centre is double the angle at the remaining circle.
Reflex ∠AOC = 2 × ∠ABC (if point B and centre are on same side of AC, but usually it refers to minor angle AOC).
∠AOC = 2 × ∠ABC = 2 × 75° = 150°.
Therefore, ∠AOC = 150°.

A line overlapping both inner and outer circles is a shared secant. The perpendicular from centre O to AD bisects both chords BC and AD.
Let M be the foot of the perpendicular.
BM = CM (because M bisects BC)
AM = DM (because M bisects AD)
Now, AB = AM - BM
CD = DM - CM
Since AM = DM and BM = CM, AB = CD.
Given AB = 4 cm, therefore CD = 4 cm.

Let ABC be an isosceles triangle with AB = AC. Therefore ∠B = ∠C.
Let DE be a line parallel to BC intersecting AB at D and AC at E.
Because DE || BC, ∠ADE = ∠B and ∠AED = ∠C (corresponding angles).
Also, ∠B = ∠C, so ∠ADE = ∠AED.
Sum of interior angles on same side of transversal: ∠BDE + ∠B = 180° ⇒ ∠BDE + ∠C = 180°.
Since the opposite angles (∠BDE and ∠C) sum up to 180°, quadrilateral DBCE is cyclic.

In right ΔAOB, OA² + OB² = AB².
Since OA = OB = r, r² + r² = (10√2)²
2r² = 200 ⇒ r² = 100 ⇒ r = 10 cm.
The radius of the circle is 10 cm.

Smaller chord AB = 6 cm, half is 3 cm. Distance = 4 cm.
Radius r = √(3² + 4²) = 5 cm.
Larger chord CD = 8 cm, half is 4 cm. Distance d = ?
In right triangle, d = √(r² - 4²) = √(5² - 16) = √(25 - 16) = √9 = 3 cm.
The larger chord is at a distance of 3 cm from the centre.

Let ABCD be a rhombus and its diagonals AC and BD intersect at O.
We know the diagonals of a rhombus bisect each other at right angles.
Therefore, ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°.
Consider the circle drawn with side AB as diameter.
The angle in a semicircle is a right angle.
Since ∠AOB = 90°, the point O must lie on the circle drawn with AB as diameter.
Thus, the circle passes through the point of intersection of diagonals.

For a cyclic quadrilateral, the sum of opposite angles is 180°.
∠A + ∠C = 180° --- (1)
Given, ∠A - ∠C = 60° --- (2)
Adding (1) and (2): 2∠A = 240° ⇒ ∠A = 120°.
Substitute ∠A in (1): 120° + ∠C = 180° ⇒ ∠C = 60°.
The angles are ∠A = 120° and ∠C = 60°.

Let AB be the diameter of a circle with centre O. Let C be any point on the circle.
The arc AB is a semicircle and subtends a straight angle at the centre, i.e., ∠AOB = 180°.
The angle subtended by an arc at the centre is double the angle it subtends on the remaining circle.
So, ∠AOB = 2 × ∠ACB.
180° = 2 × ∠ACB ⇒ ∠ACB = 90°.
Hence, the angle in a semicircle is a right angle.

Let ABC be an isosceles triangle with AB = AC.
Therefore, the base angles are equal: ∠ABC = ∠ACB.
Let the base BC be extended to D on the side B and to E on the side C.
The exterior angles are ∠ABD and ∠ACE.
By linear pair axiom: ∠ABD + ∠ABC = 180° ⇒ ∠ABD = 180° - ∠ABC.
And ∠ACE + ∠ACB = 180° ⇒ ∠ACE = 180° - ∠ACB.
Since ∠ABC = ∠ACB, we get: 180° - ∠ABC = 180° - ∠ACB.
Thus, ∠ABD = ∠ACE.
The exterior angles are equal.