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**Exercise 10.1****1. Fill in the blanks: (i) The centre of a circle lies in ____________ of the circle. (exterior/ interior) (ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior/ interior) (iii) 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) Segment of a circle is the region between an arc and _____________ of the circle. (vi) A circle divides the plane, on which it lies, in _____________ parts.**

(i) The centre of a circle lies in

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in

(iii) The longest chord of a circle is a

(iv) An arc is a

(v) Segment of a circle is the region between an arc and

(vi) A circle divides the plane, on which it lies, in

**2. Write True or False: Give reasons for your answers. (i) Line segment joining the centre to any point on the circle is a radius of the circle. (ii) A circle has only finite number of equal chords. (iii) If a circle is divided into three equal arcs, each is a major arc. (iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle. (v) Sector is the region between the chord and its corresponding arc. (vi) A circle is a plane figure.**

(i) True.

All the line segment from the centre to the circle is of equal length.

(ii) False.

We can draw infinite numbers of equal chords.

(iii) False.

We get major and minor arcs for unequal arcs. So, for equal arcs on circle we can't say it is major arc or minor arc.

(iv) True.

A chord which is twice as long as radius must pass through the centre of the circle and is diameter to the circle.

(v) False.

Sector is the region between the arc and the two radii of the circle.

(vi) True.

A circle can be drawn on the plane.

A circle is a collection of points whose every every point is equidistant from the centre. Thus, two circles can only be congruent when they the distance of every point of the both circle is equal from the centre.

Given,

AB = CD (Equal chords)

To prove,

âˆ AOB = âˆ COD

Proof,

In Î”AOB and Î”COD,

OA = OC (Radii)

OB = OD (Radii)

AB = CD (Given)

âˆ´ Î”AOB â‰… Î”COD (SSS congruence condition)

Thus, âˆ AOB = âˆ COD by CPCT.

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. Answer**

Given,

âˆ AOB = âˆ COD (Equal angles)

To prove,

AB = CD

Proof,

In Î”AOB and Î”COD,

OA = OC (Radii)

âˆ AOB = âˆ COD (Given)

OB = OD (Radii)

âˆ´ Î”AOB â‰… Î”COD (SAS congruence condition)

Thus, AB = CD by CPCT.

If chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer

No point is common.

One point P is common.

One point P is common.

Two points P and Q are common.

No point is common.**2. Suppose you are given a circle. Give a construction to find its centre. Answer**

Steps of construction:

Step I: A circle is drawn.

Step II: Two chords AB and CD are drawn.

Step III: Perpendicular bisector of the chords AB and CD are drawn.

Step IV: Let these two perpendicular bisector meet at a point. The point of intersection of these two perpendicular bisector is the centre of the circle.**3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord. Answer**

Given,

Two circles which intersect each other at P and Q.

To prove,

OO' is perpendicular bisector of PQ.

Proof,

In Î”POO' and Î”QOO',

OP = OQ (Radii)

OO' = OO' (Common)

O'P = OQ (Radii)

âˆ´ Î”POO' â‰… Î”QOO' (SSS congruence condition)

Thus,

âˆ POO' = âˆ QOO' --- (i)

In Î”POR and Î”QOR,

OP = OQ (Radii)

âˆ POR = âˆ QOR (from i)

OR = OR (Common)

âˆ´ Î”POR â‰… Î”QOR (SAS congruence condition)

Thus,

âˆ PRO = âˆ QRO

also,

âˆ PRO + âˆ QRO = 180Â°

â‡’ âˆ PRO = âˆ QRO = 180Â°/2 = 90Â°

Hence,

OO' is perpendicular bisector of PQ.**Exercise 10.4****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.****Answer**

OP = 5cm, PS = 3cm and OS = 4cm.

also, PQ = 2PR

Let RS be x.

In Î”POR,

OP^{2 }= OR^{2 }+^{ }PR^{2}

â‡’ 5^{2 }= (4-x)^{2 }+^{ }PR^{2}

â‡’ 25 = 16 + x^{2 }- 8x +^{ }^{ }PR^{2}

â‡’ PR^{2} = 9 - x^{2} + 8x --- (i)

In Î”PRS,

PS^{2 }= PR^{2 }+^{ }RS^{2}

â‡’ 3^{2 }= PR^{2} +^{ }x^{2}

â‡’ ^{ }PR^{2} = 9 - x^{2} --- (ii)

Equating (i) and (ii),

9 - x^{2} + 8x = 9 - x^{2}

â‡’ 8x = 0

â‡’ x = 0

Putting the value of x in (i) we get,

PR^{2} = 9 - 0^{2}

â‡’ PR = 3cm

Length of the cord PQ = 2PR = 2Ã—3 = 6cm**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.****Answer**

Given,

AB and CD are chords intersecting at E.

AB = CD

To prove,

AE = DE and CE = BE

Construction,

OM âŠ¥ AB and ON âŠ¥ CD. OE is joined.

Proof,

OM bisects AB (OM âŠ¥ AB)

ON bisects CD (ON âŠ¥ CD)

As AB = CD thus,

AM = ND --- (i)

and MB = CN --- (ii)

In Î”OME and Î”ONE,

âˆ OME = âˆ ONE (Perpendiculars)

OE = OE (Common)

OM = ON (AB = CD and thus equidistant from the centre)

Î”OME â‰… Î”ONE by RHS congruence condition.

ME = EN by CPCT --- (iii)

From (i) and (ii) we get,

AM + ME = ND + EN

â‡’ AE = ED

From (ii) and (iii) we get,

MB - ME = CN - EN

â‡’ EB = CE**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.****Answer**

Given,

AB and CD are chords intersecting at E.

AB = CD, PQ is the diameter.

To prove,

âˆ BEQ = âˆ CEQ

Construction,

OM âŠ¥ AB and ON âŠ¥ CD. OE is joined.

In Î”OEM and Î”OEN,

OM = ON (Equal chords are equidistant from the centre)

OE = OE (Common)

âˆ OME = âˆ ONE (Perpendicular)

Î”OEM â‰… Î”OEN by RHS congruence condition.

Thus,

âˆ MEO = âˆ NEO by CPCT

â‡’ âˆ BEQ = âˆ CEQ**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. 10.25). Answer**

OM âŠ¥ AD is drawn from O.

OM bisects AD as OM âŠ¥ AD.

â‡’ AM = MD --- (i)

also, OM bisects BC as OM âŠ¥ BC.

â‡’ BM = MC --- (ii)

From (i) and (ii),

AM - BM = MD - MC

â‡’ AB = CD**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? Answer**

Let A, B and C represent the positions of Reshma, Salma and Mandip respectively.

AB = 6cm and BC = 6cm.

Radius OA = 5cm

BM âŠ¥ AC is drawn.

ABC is an isosceles triangle as AB = BC, M is mid-point of AC. BM is perpendicular bisector of AC and thus it passes through the centre of the circle.

Let AM = y and OM = x then BM = (5-x).

Applying Pythagoras theorem in Î”OAM,

OA^{2 }= OM^{2 }+^{ }AM^{2}

â‡’ 5^{2 }= x^{2 }+^{ }y^{2 }--- (i)

Applying Pythagoras theorem in Î”AMB,

AB^{2 }= BM^{2 }+^{ }AM^{2}

â‡’ 6^{2 }= (5-x)^{2 }+^{ }y^{2} --- (ii)

Subtracting (i) from (ii), we get

36 - 25 = (5-x)^{2 }-^{ }x^{2} -

â‡’ 11 = 25 - 10x

â‡’ 10x = 14 â‡’ x= 7/5

Substituting the value of x in (i), we get

y^{2 }+ 49/25 = 25

â‡’ y^{2} = 25 - 49/25

â‡’ y^{2} = (625 - 49)/25

â‡’ y^{2} = 576/25

â‡’ y = 24/5

Thus,

AC = 2Ã—AM = 2Ã—y = 2Ã—(24/5) m = 48/5 m = 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. Answer**

Let A, B and C represent the positions of Ankur, Syed and David respectively. All three boys at equal distances thus ABC is an equilateral triangle.

AD âŠ¥ BC is drawn. Now, AD is median of Î”ABC and it passes through the centre O.

Also, O is the centroid of the Î”ABC. OA is the radius of the triangle.

OA = 2/3 AD

Let the side of a triangle a metres then BD = a/2 m.

Applying Pythagoras theorem in Î”ABD,

AB^{2 }= BD^{2 }+^{ }AD^{2}

â‡’ AD^{2 }= AB^{2 }- BD^{2 }

â‡’ AD^{2 }= a^{2 }-^{ }(a/2)^{2}

â‡’ AD^{2 }= 3a^{2}/4

â‡’ AD = âˆš3a/2

OA = 2/3 AD â‡’ 20 m = 2/3 Ã— âˆš3a/2

â‡’ a = 20âˆš3 m

Length of the string is 20âˆš3 m.