Class Nine Maths NCERT Solution Chapter 10 Circle
Chapter 10: Circle
Exercise 10.1
Q 1: Fill in the blanks:
- The centre of a circle lies in …………..of the circle. (exterior/ interior)
- A point, whose distance from the centre of a circle is greater than its radius lies in …………of the circle. (exterior/ interior)
- The longest chord of a circle is a ………..of the circle.
Iv. An arc is a ………….when its ends are the ends of a diameter.
v. A segment of a circle is the region between an arc and ……………of the circle.
v. A circle divides the plane, on which it lies, in …………….parts.
Sol: (i) interior, (ii) exterior, (iii) diameter, (iv) semicircle, (v) chord, (vi) three
Q 2: Write True or False
(i) The line segment joining the centre to any point on the circle is a radius of the circle.
Sol: True; because the distance between the centre and any point on the circumference is called radius.
(ii) A circle has an only a finite number of equal chords.
Sol: False, because a circle can have an infinite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
Sol: False, because a major arc is always bigger than a minor arc, and they cannot be equal.
(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
Sol: True because the diameter is the longest chord of a circle and is always twice the radius.
(v) The sector is the region between the chord and its corresponding arc.
Sol: False, the region between the chord and its corresponding arc is called a segment.
(vi) A circle is a plane figure.
Sol: True
Exercise 10.2
Q1: 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.
Sol: A per the theorem; equal chords (of a circle) subtend equal angles at the centre. Hence, it is clear that equal chords of congruent circles would subtend equal angles at their centres.
Q 2: Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Sol: This is same as the previous question. The theorem says that if two chords subtend equal angles at the centre then the chords are equal. Hence, if two chords in two congruent circles subtend equal angles at their centres then the chords are equal.
Exercise 10.3
Q 1: Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Sol: The maximum number of common points between a pair of circles can be two. Thus, it can also be zero or 1.
Q2: Suppose you are given a circle. Give the construction to find its centre.
Sol: Here, an arc PQ is given; as shown in the figure.
Take a point R on the arc.
Join the point PR and PQ.
Draw perpendicular bisectors of PR and PQ.
It is observed that perpendicular bisectors of chords PR and PQ intersect at O.
Taking O as centre and OR as radius, draw the circle.
You will find that the circle is passing through P, R and Q.
This proves that through any three non-collinear points; one and only one circle can pass.
Q 3: If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Sol: According to question, two circles intersect at points A and B.
To Prove: AM = MB
∠AMO = ∠BMO = 90°
In triangles AMO and
BMO;
AO = BO (radii of circle)
OM = OM (Since common side)
So, from SSS theorem; ΔAMO ≈ ΔBMO
Thus, AM = MB proved.
Hence, ∠AMO = ∠BMO = 90° proved.
Thus, it is proved that when two circles intersect at two points then their centres lie on the perpendicular bisector of the common chord.