Class 9 Mathematics
Chapter 9: Circles
Exercise 9.2 – Stepwise Solutions
Important Concepts Used
1. The line joining the centres of two intersecting circles is perpendicular to the common chord.
2. Equal chords of the same circle are equidistant from the centre.
3. The perpendicular from the centre of a circle to a chord bisects the chord.
4. In the same circle, equal chords subtend equal angles at the centre.
5. Chords equidistant from the centre are equal.
Q1. 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.
Given:
Radius of first circle = 5 cm
Radius of second circle = 3 cm
Distance between centres = 4 cm
Let the centres be O and O′, and let AB be the common chord.
The line joining the centres is perpendicular to the common chord.
Let M be the mid-point of chord AB.
Then, in right triangles OMA and O′MA,
OA = 5 cm, O′A = 3 cm, and OO′ = 4 cm
Let OM = x, then O′M = 4 − x
Also, MA is common for both triangles.
Using Pythagoras theorem:
x² + MA² = 25
(4 − x)² + MA² = 9
Subtracting,
x² − (4 − x)² = 25 − 9
x² − (16 − 8x + x²) = 16
8x − 16 = 16
8x = 32
x = 4
So, O′M = 4 − 4 = 0
That means M coincides with the centre of the smaller circle.
Hence the common chord passes through the centre of the smaller circle, so it is its diameter.
Therefore,
Length of common chord = 2 × 3 = 6 cm
Q2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to the corresponding segments of the other chord.
Let equal chords AB and CD intersect at E inside the circle.
We have to prove that:
AE = DE and BE = CE
Given:
AB = CD
Since E is the point of intersection,
AB = AE + EB
CD = CE + ED
But AB = CD, so
AE + EB = CE + ED …(1)
Also, by the intersecting chords theorem,
AE × EB = CE × ED …(2)
Now two numbers with the same sum and same product are equal in corresponding order.
Therefore,
AE = DE and EB = CE
Hence proved.
Q3. 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.
Let equal chords AB and CD intersect at E, and O be the centre of the circle.
From Q2, we know:
AE = DE and BE = CE
Also, OA = OD (radii of the same circle)
OE = OE (common side)
Now in triangles OEA and OED:
OA = OD
OE = OE
AE = DE
Therefore,
ΔOEA ≅ ΔOED by SSS
So, by CPCT,
∠OEA = ∠OED
Similarly, in triangles OEB and OEC:
OB = OC
OE = OE
BE = CE
Therefore,
ΔOEB ≅ ΔOEC by SSS
So,
∠OEB = ∠OEC
Hence, the line OE makes equal angles with the two chords.
Therefore, the line joining the point of intersection to the centre makes equal angles with the chords.
Q4. If a line intersects two concentric circles with the same centre O at A, B, C and D, prove that AB = CD.
Let the line intersect the outer circle at A and D, and the inner circle at B and C, in the order A, B, C, D.
Draw OM perpendicular to the line.
Since O is the centre of both circles:
The perpendicular from the centre to a chord bisects the chord.
So, for chord AD of the outer circle,
AM = MD
And for chord BC of the inner circle,
BM = MC
Now,
AB = AM − BM
CD = DM − CM
But AM = DM and BM = CM
Therefore,
AB = CD
Hence proved.
Q5. Three girls Reshma, Salma and Mandip are standing on a circle of radius 5 m. 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 6 m each, what is the distance between Reshma and Mandip?
Let the three girls be at points R, S and M on the circle.
Given:
RS = 6 m
SM = 6 m
Since RS and SM are equal chords of the same circle, they subtend equal angles at the centre.
Equal chords of the same circle are equal.
So the chord joining Reshma and Mandip is also equal.
Therefore,
RM = 6 m
Answer: The distance between Reshma and Mandip is 6 m.
Q6. A circular park of radius 20 m 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.
The three boys are equally spaced on the boundary of the circle.
So they form an equilateral triangle inside the circle.
Radius of the circle = 20 m
Let the side of the equilateral triangle be a.
For an equilateral triangle,
Circumradius R = a / √3
So,
a = R√3
= 20√3 m
Thus, the distance between any two boys is:
20√3 m
This is the length of the string of each toy phone.
Answer: 20√3 m