Circles ⇒⇒ Exercise 9.2

Question 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.

Solution :

Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively, and the distance between the centers of two circles is 4 Cm.

Given that the circles intersect at 2 points, M and N

Let MN be the common chord.

∴ OM = 5 Cm. and O'M = 3 Cm.

It can be observed that the radius of the bigger circle is more than the distance between the 2 centres, we can say that the centre of smaller circle lies inside the bigger circle itself.

In two intersecting circles, The line segment connecting the centers of the two circles (OO') is the perpendicular bisector of the common chord MN.

∴ ML = LN and $\angle $ OLM = $\angle $ O'LM = 90°

Given that OO' = 4 Cm.

Consider that OL = x, so O'L = (4 - x)

Now, In the right-angled ∆ OLM ( using the larger circle's radius), we have

By using Pythagoras theorem

OM² = ML² + OL²

⇒ 5² = ML² + x²

⇒ ML² = 25 - x² ....(i)

Now, In the right-angled ∆ O'LM ( using the smaller circle's radius), we have

By using Pythagoras theorem

O'M² = ML² + O'L²

⇒ 3² = ML² + (4-x)²

⇒ ML² = 9 - (4-x)²

⇒ ML² = 9 - 16 + x² − 8x

⇒ ML² = - 7 - x² + 8x .....(ii)

From equation (i) and (ii)

25 - x² = - 7 - x² + 8x

⇒ 8x = 32

⇒ x = 4

Hence, O'L = 4 - 4 = 0 cm ,

L and O′ coincide .

Thus, the common chord will pass through the center of the smaller circle

∴ MN is a diameter of the smaller circle.

⇒ ML = O'M = 3 cm

( Radius of smaller circle )

⇒ Length of the common chord MN = 2 × ML

⇒ (2 × 3) cm

⇒ MN = 6 Cm.

Thus, the required length of the common chord = 6 Cm .

Question 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.

Solution :

Given: Circle with centre O, MN and XY chords are equal and intersect at D

MN = XY

To prove: XD = MD and YD = ND

Construction: Join the center O to the intersection point OD

Draw OB ⊥ XY and OA ⊥ MN .

In ∆OAD and ∆OBD,

$ \angle $ OAD = $ \angle $ OBD = 90°

( Construction perpendiculars from the center to the chords )

OD = OD

( Common )

OA = OB

( A theorem in circle geometry states that equal chords of a circle are equidistant from the center. )

Thus , ∆OAD ≅ ∆OBD

( By the Right-Hypotenuse-Side (RHS) Congruence Criterion: )

∴ AD = BD ....(i)

( By CPCT )

Now , Given that MN = XY

AM = AN = BX = BY .....(ii)

( ∵ Perpendicular from the centre to the chord bisects the chord. )

We know that:

MD = AM + AD and

XD = BX + BD

From equation (i) and (ii)

⇒ XD = AM + AD

⇒ ∴ it is evident MD = XD

Now,

⇒ ND = MN - MD and

⇒ YD = XY - XD

MD and XD are equal, MN and XY are equal. So ND and YD are also equal.

ND = YD

Hence proved,
Therefore, the segments of one chord are equal to the corresponding segments of the other chord.

Question 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.

Solution :

Given: Circle with centre O, MN and XY chords are equal and intersect at D

MN = XY

To prove: $ \angle $ ODA = $ \angle $ ODB

Construction: Join OD

Draw OB ⊥ XY and OA ⊥ MN .

In ∆OAD and ∆OBD,

$ \angle $ OAD = $ \angle $ OBD = 90°

( Construction perpendiculars from the center to the chords )

OD = OD

( Common )

OA = OB

( A theorem in circle geometry states that equal chords of a circle are equidistant from the center. )

Thus , ∆OAD ≅ ∆OBD

( By the Right-Hypotenuse-Side (RHS) Congruence Criterion: )

∴ $ \angle $ ODA = $ \angle $ ODB

( By CPCT )

Hence proved,

Question 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.

Solution :

Given: A line intersects two concentric circles with centre O at A, B, C, and D.

To prove: AB = CD

Construction: Draw OM ⊥ AD

We can see that BC is the chord of the inner circle

OM ⊥ BC

(∵ A theorem in circle geometry states that the perpendicular drawn from the center of a circle to a chord bisects the chord. )

∴ BM = MC ... (i)

Similarly, We can see that AD is the chord of the outer bigger circle

OM ⊥ AD

( ∵ Perpendicular from the centre to the chord bisects the chord. )

∴ AM = MD ... (ii)

Subtracting equation (i) From (ii),we obtain

AM – BM = MD – MC

∴ AB = CD

Therefore, it is proven that the segments of the line cut off between the two concentric circles are equal.

Question 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 6 m each, what is the distance between Reshma and Mandip ?

Solution :

Given: The circle of radius 5 m.

Let O be the centre of the circle and Reshma, Salma and Mandip be positioned at R, S and M respectively.

RS = SM = 6 m

Since RS = SM = 6 m, the chords are equal. This means △RSM is an isosceles triangle.

Also, since OR = OS = OM = 5 m, the triangle △ORM is also isosceles.

To prove: Distance between Reshma and Mandip , RM = ?

Construction: Draw perpendicular OX on RS . Join OS

We can see that RS is the chord of the circle

OX ⊥ RS

( ∵ Perpendicular from the centre to the chord bisects the chord )

∴ RX = SX = 3 m

Now, in Δ ORX

Using Pythagoras theorem we get,

OR² = OX² + RX²

⇒ (5)² = (OX)² + (3)²

⇒ (OX)² = 25 -9

⇒ (OX)² = 16

⇒ OX = 4 m

Now , We know that

Area of triangle = 1/2 × B × H

Area of ΔORS = 1/2 × OS × RY

⇒ 1/2 × 5 × RY .....(i)

( ∵ OS = 5 m Radii of circle )

Now , Again Area of ΔORS

Area of ΔORS = 1/2 × OX × RS

⇒ 1/2 × 4 × 6 .....(i)

Area of ΔORS = 12 m .....(ii)

From equation (i) and (ii),we obtain

1/2 × 5 × RY = 12

RY × 5 = 12 × 2

RY = 24 / 5

RY = 4.8 m

As we know the perpendicular from the centre of a circle bisects the chord.

OY ⊥ RM

∴ RM = 2RY

RM = 2 × 4.8

RM = 9.6 m

The distance between Reshma and Mandip is , RM = 9.6 meters.

Question 6

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 ?

Solution :

Given: The circular park of radius 20 m.

Let O be the centre of the circle and Ankur, Syed and David be positioned at A, S and D respectively.

OA = OS = OD = 20 m

To prove: Length of wire of telephone = ?

According to the given question, AS = AD = SD

∴ ASD is an equilateral triangle

Let AS = AD = SD = 2x

We can see that SD is the chord of the circle

Construction: Draw perpendicular OP on SD .

OP ⊥ SD

( ∵ Perpendicular from the centre to the chord bisects the chord )

∴ SP = DP = x m

Now, in Δ APD

Using Pythagoras theorem we get,

AD² = AP² + DP²

⇒ (2x)² = (AP)² + x²

⇒ (AP)² = 4x² - x²

⇒ (AP)² = 3x²

⇒ AP = √3x .....(i)

Now , We know that

OP = AP - AO

⇒ OP = √3x - 20 .....(ii)

Now, in Δ OPD

Using Pythagoras theorem we get,

OD² = OP² + DP²

⇒ (20)² = (√3x - 20)² + x²

⇒ 400 = 3x² -40√3x + 400 + x²

⇒ 4x² -40√3x = 0

⇒ 4x(x -10√3) = 0

Since x is not equal to 0 , then

x - 10√3 = 0

x = 10√3

Now,

SD = 2 × SP

SD = 2 × 10√3 = 20√3 m

Hence , the length of each string = 20√3 m