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A circle is a collection of all the points that are at a distance from a particular point in a fixed plane. It is a closed curve in which all its points are at an equal distance from its centre. Its fixed point is the centre and the equal distance of the points in the curve to the centre is called the radius. A circle’s equal chords tend to subtend equal angles at the centre. The perpendicular to a chord from the centre of a circle bisects the chord.
The line that is drawn through the circle’s centre bisecting a chord tends to be perpendicular to that chord.
Only one circle can pass through three non-collinear points. Chords that are equal are at equal distances from the centre in a circle. The angle that is subtended the centre by an arc is two times the angle that is subtended by it at a point on the circle’s remaining point. Angles that are found in a circle’s same segment are equal. The sum of opposite angles in a cyclic quadrilateral is 180°.
Ques1 : What part of the congruent circles subtend equal angles at the centres.
- Radii
- Segments
- Arcs
- Chords
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Ans: (d) Chords
Explanation:
Let the ΔCOD and ΔAOB be two triangles inside of the circle.
As we know,
OB = OD and OA = OC (As they are radii of the circle)
AB = CD (Given)
So, by SSS congruency ΔCOD ≅ ΔAOB.
Hence, By the CPCT rule, ∠COD = ∠AOB.
Hence, this proves the statement that equal chords of the congruent circles subtend equal angles at the centres.
Ques2 : If chords CD and AB of two congruent circles are subtending equal angles at their centres, then:
- AB > CD
- AB = CD
- None of the above
- AB < AD
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Answer: (b) AB = CD
Explanation: Upon noticing the figure in the question mentioned above.
In triangles COD and AOB,
Given, ∠AOB = ∠COD
OB = OD and OA = OC (these are the radii of the circle)
So, ΔCOD ≅ ΔAOB. (by SAS congruency)
Hence, by CPCT, CD = AB (By CPCT)
Ques3 : The diameter of a semi-circle subtends an angle of how many degrees?
- 45
- 90
- 60
- 180
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Ans: (d) 180
Explanation: The semicircle is a shape that is made from half of a circle, hence a semicircle’s diameter will be a straight line that subtends 180 degrees.
Ques4 : In the given figure. CD and AB are two chords of a circle that intersect at point E. Then:
- ∠BEQ = ∠CEQ
- ∠BEQ > ∠CEQ
- None of the above
- ∠BEQ < ∠CEQ
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Ans: (a) ∠BEQ = ∠CEQ
Explanation: As we know that the equal chords always tend to be equidistant from the centre OM = ON.
OE = OE (Common side)
∠ONE = ∠OME (both are perpendiculars)
So, ΔOEN ≅ ΔOEM (by RHS similarity)
Therefore, ∠NEO = ∠MEO (by CPCT)
Hence, ∠CEQ = ∠BEQ
Ques5 : If two concentric circles with centre O at A, B, C and D are intersected by a line, then:
- AB > CD
- AB = CD
- None of the above
- AB < CD
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Answer: (b) AB = CD
Explanation:
From the above fig, we know that, OM ⊥ AD.
Therefore, MD = AM → 1
Also, as OM ⊥ BC, OM will bisect BC.
Therefore, MC = BM → 2
Subtracting eq 2 from 1, we get,
AM – BM = MD – MC
Hence, AB = CD
Ques6 : Find the value of the ∠ADC in the figure given below:
- 30°
- 60°
- 55°
- 45°
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Answer: (d) 45°
Explanation: As we observed, ∠AOC = ∠AOB + ∠BOC
So, ∠AOC = 30° + 60°
So, ∠AOC = 90°
An angle that is subtended by the circle’s arc at the centre is two times the angle subtended by that arc at any point of the remaining circle.
So, ∠ADC = ½ of ∠AOC
= ½ × 90° = 45°
Ques7 : Find the angle OPR in the figure given below.
- 15°
- 20°
- 10°
- 12°
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Answer: (c) 10°
Explanation: The angle that is subtended at the centre of a circle by an arc is two times the angle subtended by that arc at any point on the remaining circle.
Hence, ∠POR = 2 × ∠PQR
As we know that angle PQR is 100°
So, ∠POR = 2 × 100° = 200°
Hence, ∠POR = 360° – 200° = 160°
Now, in ΔOPR,
OR and OP are the radii of a circle.
So, OR = OP
Also, ∠ORP = ∠OPR
By the angle sum property of the triangle, we know that
∠OPR + ∠POR + ∠ORP = 180°
∠OPR + ∠OPR = 180° – 160°
As, ∠ORP = ∠OPR
2 x ∠OPR = 20°
Therefore, ∠OPR = 10°
Ques8 : Find the ∠CAO In the figure given below if ∠AOB = 90º and ∠ABC = 30º.
- 45º
- 30º
- 90º
- 60º
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Answer: (d) 60º
Explanation: Given: ∠ABC = 30º and ∠AOB = 90º.
OB = OA (These are the Radii of the circle)
Let x = ∠OAB = ∠OBA
So, In the triangle OAB,
∠OBA + ∠OAB + ∠AOB = 180° (according to the angle sum property of triangle)
⇒ x + x + 90° = 180°
⇒ 2x = 180° – 90°
⇒ x = 90°/2
= 45°
Therefore, We know that ∠OBA = 45° and ∠OAB = 45°.
Now as we know The angle that is subtended at the centre of a circle by an arc is two times the angle subtended by that arc at any point on the remaining circle, we can write
∠AOB = 2 x ∠ACB
This can be rewritten as,
∠ACB = ½ ∠AOB = (½) × 90° = 45°
Now, by applying the angle sum property of triangle on the triangle ABC, we get,
∠BAC + ∠ACB + ∠CBA = 180°
[∠BAO + ∠CAO] + ∠CBA + ∠ACB = 180° (As, ∠BAC = ∠CAO + ∠BAO)
Now, by substituting the values, we get
45° + (45° + ∠CAO) + 30° = 180°
∠CAO = 180°- (30° + 45° + 45°)
∠CAO = 180°-120°
∠CAO = 60°
Therefore, ∠CAO is equal to 60°.
Ques9 : Find the ∠BAC in the cyclic quadrilateral ABCD in which AB is a circle’s diameter that has circumscribed it and ∠ADC = 140º.
- 40º
- 30º
- 80º
- 50º
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Answer: (d) 50º
Explanation: The Given quadrilateral ABCD is cyclic and ∠ADC = 140º.
As we know that the sum of opposite angles of a cyclic quadrilateral is 180°.
So, ∠ABC + ∠ADC = 180°
Now, by substituting ∠ADC = 140º in the above equation, we get
140° + ∠ABC = 180°
∠ABC = 180° – 140° = 40°
since the angle that is subtended by the diameter at the circle’s circumference is 90°.
We get, ∠ACB = 90°
Now, By using the angle property of triangle in the triangle, ABC,
∠ABC + ∠CAB + ∠ACB = 180°
∠CAB + 90° + 40° + = 180°
∠CAB = 180° – 90° – 40°
∠CAB = 50°
Therefore, ∠BAC or ∠CAB = 50°.
Ques 10: In the figure given below, find the ∠ACB, if ∠OAB = 40º.
- 50º
- 40º
- 70º
- 60º
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Answer: (a) 50º
Explanation: Given: ∠OAB = 40º,
In the triangle OAB,
OB = OA (both are radii)
Since, the angles that are opposite to the equal sides tend to be equal,
∠OBA = ∠OAB (i.e.) ∠OBA = 40°
Now, we use the angle sum property of triangle to write
∠AOB + ∠OBA + ∠BAO = 180°
Now, by substituting the values we get,
∠AOB + 40° + 40° = 180°
∠AOB = 180 – 80° = 100°
∠AOB = 2 x ∠ACB ∠ACB = ½ ∠AOB
Hence, ∠ACB = 100°/2 = 50°.
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