CIRCLES - Q&A
Exercise 9.1
1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Given: Two congruent circles with centres O and O' and radii r. AB is a chord of the first circle and CD is a chord of the second circle such that AB = CD.
To Prove: ∠ AOB = ∠ CO'D.
Proof:
In Δ AOB and Δ CO'D:
1. AB = CD (Given)
2. OA = O'C (Radii of congruent circles)
3. OB = O'D (Radii of congruent circles)
Therefore, by SSS congruence rule:
Δ AOB ≅ Δ CO'D.
By CPCT (Corresponding Parts of Congruent Triangles):
∠ AOB = ∠ CO'D.
Hence, equal chords of congruent circles subtend equal angles at their centres.
2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Given: Two congruent circles with centres O and O'. ∠ AOB = ∠ CO'D.
To Prove: Chord AB = Chord CD.
Proof:
In Δ AOB and Δ CO'D:
1. OA = O'C (Radii of congruent circles)
2. ∠ AOB = ∠ CO'D (Given)
3. OB = O'D (Radii of congruent circles)
Therefore, by SAS congruence rule:
Δ AOB ≅ Δ CO'D.
By CPCT:
AB = CD.
Hence, the chords are equal.
Exercise 9.2
1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Let the two circles be with centres O and O' and radii 5 cm and 3 cm respectively.
Let them intersect at points A and B. So AB is the common chord.
The distance between centres OO' = 4 cm.
Let OO' intersect AB at M.
We know that the line joining centres is the perpendicular bisector of the common chord.
So, AB ⊥ OO' and M is the mid-point of AB.
Let OM = x cm. Then O'M = (4 - x) cm.
In right-angled Δ OMA:
OA2 = OM2 + AM2
52 = x2 + AM2
AM2 = 25 - x2 ... (i)
In right-angled Δ O'MA:
O'A2 = O'M2 + AM2
32 = (4 - x)2 + AM2
AM2 = 9 - (4 - x)2 ... (ii)
From (i) and (ii):
25 - x2 = 9 - (16 + x2 - 8x)
25 - x2 = 9 - 16 - x2 + 8x
25 = -7 + 8x
32 = 8x ⇒ x = 4.
Since x = 4, the centre O' coincides with M (because OO' = 4).
Substitute x = 4 in (i):
AM2 = 25 - 42 = 25 - 16 = 9.
AM = 3 cm.
Length of common chord AB = 2 × AM = 2 × 3 = 6 cm.
(Note: Since radius of smaller circle is 3 cm and distance between centres is 4 cm and radius of larger is 5 cm, the common chord passes through the centre of the smaller circle).
2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Given: A circle with centre O. Equal chords AB and CD intersect at P.
To Prove: AP = CP and BP = DP.
Construction: Draw OM ⊥ AB and ON ⊥ CD. Join OP.
Proof:
In right-angled Δ OMP and Δ ONP:
1. OP = OP (Common hypotenuse)
2. OM = ON (Equal chords are equidistant from centre)
3. ∠ OMP = ∠ ONP = 90°
So, Δ OMP ≅ Δ ONP (by RHS).
Therefore, MP = NP (by CPCT).
Since AB = CD, their halves are also equal.
AM = CN ... (i)
Also, MP = NP ... (ii)
Adding (i) and (ii):
AM + MP = CN + NP
AP = CP.
Subtracting AP = CP from AB = CD:
AB - AP = CD - CP
BP = DP.
Hence Proved.
3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Given: Equal chords AB and CD intersect at P. O is the centre.
To Prove: ∠ OPA = ∠ OPC (or ∠ OPE = ∠ OPF if we take perpendiculars).
Construction: Draw OM ⊥ AB and ON ⊥ CD. Join OP.
Proof:
In right-angled Δ OMP and Δ ONP:
1. OP = OP (Common)
2. OM = ON (Equal chords are equidistant from centre)
3. ∠ OMP = ∠ ONP = 90°
So, Δ OMP ≅ Δ ONP (by RHS).
Therefore, ∠ OPM = ∠ OPN (by CPCT).
This means the line OP makes equal angles with the chords AB and CD.
4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 9.12).
Given: A line intersects two concentric circles at A, B, C, D.
To Prove: AB = CD.
Construction: Draw OM ⊥ AD.
Proof:
For the outer circle, AD is a chord and OM ⊥ AD.
Perpendicular from centre bisects the chord.
So, AM = DM ... (i)
For the inner circle, BC is a chord and OM ⊥ BC.
So, BM = CM ... (ii)
Subtracting (ii) from (i):
AM - BM = DM - CM
AB = CD.
Hence Proved.
5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?
Let positions of Reshma, Salma, Mandip be R, S, M.
RS = 6m, SM = 6m. Radius OS = 5m, OR = 5m, OM = 5m.
Since RS = SM, quadrilateral ORSM is a kite-like structure (adjacent sides equal).
But actually, R, S, M are on a circle.
Since RS = SM, equal chords subtend equal angles at centre. ∠ ROS = ∠ SOM.
So, OS bisects ∠ ROM. Therefore OS ⊥ RM and bisects RM.
Let OS intersect RM at K.
In Δ ORS, sides are 5, 5, 6.
Semi-perimeter s = (5+5+6)/2 = 8.
Area of Δ ORS = √[s(s-a)(s-b)(s-c)] = √[8(3)(3)(2)] = √144 = 12 m2.
Also Area of Δ ORS = 1/2 × Base × Height = 1/2 × OS × RK.
12 = 1/2 × 5 × RK
24 = 5 × RK
RK = 4.8 m.
Since OS bisects RM, RM = 2 × RK.
RM = 2 × 4.8 = 9.6 m.
Distance between Reshma and Mandip is 9.6 m.
6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
Let Ankur, Syed, David be at A, S, D.
Since they are at equal distances, AS = SD = DA.
So Δ ASD is an equilateral triangle. Let side be 'a'.
Radius circumscribing the equilateral triangle R = 20m.
We know relation between side of equilateral triangle and circumradius:
R = a / √3
20 = a / √3
a = 20√3 m.
So, the length of the string of each phone is 20√3 m.
Exercise 9.3
1. In Fig. 9.23, A, B and C are three points on a circle with centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC.
∠ AOC = ∠ AOB + ∠ BOC
∠ AOC = 60° + 30° = 90°.
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
So, ∠ AOC = 2 × ∠ ADC.
90° = 2 × ∠ ADC
∠ ADC = 45°.
2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Let chord be AB and centre be O. Radius OA = OB = r.
Given Chord AB = r.
So, Δ OAB is an equilateral triangle.
Therefore, angle at centre ∠ AOB = 60°.
Angle at major arc (say at point P):
∠ APB = 1/2 ∠ AOB = 1/2 (60°) = 30°.
Angle at minor arc (say at point Q):
AQBP is a cyclic quadrilateral.
∠ APB + ∠ AQB = 180° (Opposite angles sum to 180°)
30° + ∠ AQB = 180°
∠ AQB = 150°.
Answer: Angle at major arc = 30°, Angle at minor arc = 150°.
3. In Fig. 9.24, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR.
Take a point S on the major arc.
PQRS is a cyclic quadrilateral.
∠ PQR + ∠ PSR = 180°
100° + ∠ PSR = 180°
∠ PSR = 80°.
Angle at centre ∠ POR = 2 × ∠ PSR = 2 × 80° = 160°.
In Δ OPR, OP = OR (Radii).
So, ∠ OPR = ∠ ORP.
Sum of angles = 180°
∠ OPR + ∠ ORP + ∠ POR = 180°
2 ∠ OPR + 160° = 180°
2 ∠ OPR = 20°
∠ OPR = 10°.
4. In Fig. 9.25, ∠ ABC = 69°, ∠ ACB = 31°, find ∠ BDC.
In Δ ABC:
∠ BAC + ∠ ABC + ∠ ACB = 180°
∠ BAC + 69° + 31° = 180°
∠ BAC + 100° = 180°
∠ BAC = 80°.
Angles in the same segment are equal.
Therefore, ∠ BDC = ∠ BAC.
∠ BDC = 80°.
5. In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.
Step 1: Find ∠ CED and ∠ EDC.
∠ CED + ∠ BEC = 180° (Linear Pair)
∠ CED + 130° = 180°
∠ CED = 50°.
In Δ CDE:
∠ CDE + ∠ DCE + ∠ CED = 180°
∠ CDE + 20° + 50° = 180°
∠ CDE = 110°.
(Note: ∠ CDE is same as ∠ CDB).
Step 2: Find ∠ BAC.
Angles in the same segment are equal.
∠ BAC = ∠ BDC (Subtended by arc BC).
Since ∠ BDC = ∠ CDE = 110°.
So, ∠ BAC = 110°.
6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°, ∠ BAC is 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD.
Step 1: Find ∠ BDC.
Angles in the same segment are equal.
∠ BDC = ∠ BAC.
Since ∠ BAC = 30°, ∠ BDC = 30°.
Step 2: Find ∠ BCD.
In Δ BCD:
∠ DBC + ∠ BDC + ∠ BCD = 180°
70° + 30° + ∠ BCD = 180°
100° + ∠ BCD = 180°
∠ BCD = 80°.
Step 3: Find ∠ ECD if AB = BC.
If AB = BC, then in Δ ABC, ∠ BAC = ∠ BCA.
So, ∠ BCA = 30°.
We know ∠ BCD = 80°.
∠ BCD = ∠ BCA + ∠ ACD (or ∠ ECD)
80° = 30° + ∠ ECD
∠ ECD = 50°.
7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Let ABCD be the cyclic quadrilateral and AC, BD be diagonals which are diameters.
Since AC is a diameter, angle in a semi-circle is 90°.
So, ∠ ABC = 90° and ∠ ADC = 90°.
Since BD is a diameter:
∠ BAD = 90° and ∠ BCD = 90°.
Since all angles of the quadrilateral are 90°, ABCD is a rectangle.
8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Given: Trapezium ABCD with AB || CD and AD = BC.
To Prove: ABCD is cyclic (i.e., ∠ A + ∠ C = 180°).
Construction: Draw DE ⊥ AB and CF ⊥ AB.
Proof:
In right Δ DEA and Δ CFB:
1. AD = BC (Given)
2. DE = CF (Distance between parallel lines)
3. ∠ DEA = ∠ CFB = 90°
So, Δ DEA ≅ Δ CFB (by RHS).
Therefore, ∠ A = ∠ B (by CPCT).
Since AB || CD, consecutive interior angles sum to 180°.
∠ A + ∠ D = 180°
∠ B + ∠ C = 180°
Substitute ∠ B with ∠ A in the second equation:
∠ A + ∠ C = 180°.
Since opposite angles sum to 180°, ABCD is a cyclic quadrilateral.
9. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 9.27). Prove that ∠ ACP = ∠ QCD.
Proof:
For the first circle (left side):
Angles in the same segment are equal.
∠ ACP = ∠ ABP (Subtended by arc AP). ... (i)
For the second circle (right side):
∠ QCD = ∠ QBD (Subtended by arc QD). ... (ii)
But ∠ ABP = ∠ QBD (Vertically opposite angles). ... (iii)
From (i), (ii), and (iii):
∠ ACP = ∠ QCD.
Hence Proved.
10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.
Given: Δ ABC. Two circles with diameters AB and AC intersect at A and D.
To Prove: D lies on BC.
Proof:
Join AD.
Since AB is a diameter, ∠ ADB is an angle in a semicircle.
So, ∠ ADB = 90°.
Since AC is a diameter, ∠ ADC is an angle in a semicircle.
So, ∠ ADC = 90°.
Adding angles:
∠ ADB + ∠ ADC = 90° + 90° = 180°.
This implies that BDC is a straight line.
Therefore, D lies on the line segment BC.
11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠ CBD.
Given: ∠ ABC = 90° and ∠ ADC = 90°.
Proof:
Since ∠ ABC + ∠ ADC = 90° + 90° = 180°, the quadrilateral ABCD is cyclic (Sum of opposite angles is 180°).
Consider the circle passing through A, B, C, D.
Angles in the same segment are equal.
∠ CAD and ∠ CBD are in the same segment (subtended by arc CD).
Therefore, ∠ CAD = ∠ CBD.
Hence Proved.
12. Prove that a cyclic parallelogram is a rectangle.
Given: ABCD is a cyclic parallelogram.
To Prove: ABCD is a rectangle.
Proof:
Since ABCD is a parallelogram, opposite angles are equal.
∠ A = ∠ C.
Since ABCD is cyclic, sum of opposite angles is 180°.
∠ A + ∠ C = 180°.
Substitute ∠ C with ∠ A:
2 ∠ A = 180°
∠ A = 90°.
Since a parallelogram with one angle 90° is a rectangle, ABCD is a rectangle.