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# Class 9 NCERT Solutions- Chapter 10 Circles – Exercise 10.6

• Last Updated : 25 Jan, 2021

### Question 1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Solution:

Given: Two circles with Centre A and B circle intersects at C and D.

BC=BD           ———[radii of the same circle]

AB=AB            ——–[common]

### Question 2. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Solution:

Let O be the centre of circle and be r cm

Given: AB=5cm, CD =11cm

Construction: Draw OM perpendicular AB  and  OL perpendicular CD.

Because OM perpendicular AB  and  OL perpendicular CD and AB||CD.

âˆ´Points O,L, and  M are collinear, than âˆ M=6cm

Let OL=x

Then OM=6=x

Join AO and CO

OA=OC =r

OL=1/2CD=1/2*11=5.5cm  —–[perpendicular from bisects the chord]

AM=1/2AB=1/2*5=2.3cm   —–[perpendicular from bisects the chord]

Now, In right âˆ†DLC

r2=(OL)2+(CL)2

r2=x2+(5.5)2

r2=x2+30.25       ———–1

Now in right âˆ†OMA

r2=(OM)2+(MA)2

r2=(6-x)2+(2.5)2

r2=36+x2=12x+6.25

r2=x2-12x+42.25             ———–2

Now equating equation 1 and 2

X2+30.25=x2-12x+30.25

12x=42.25-30.25

X=12/12=1

Putting value of x in equation 1

r2=x2+30.25

r2=(1)2+30.25

r2=31.25

r=âˆš31.25=5.6  (approx.)

### Question 3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the center, what is the distance of the other chord from the centre?

Solution:

Let AB and CD are|| chord of circle with centre O which AB=6cm and CD=8cm and radius of circle =r cm.

Construction: Draw OP perpendicular AB  and OM perpendicular CD.

Because AB||CD and OP perpendicular AB and OM perpendicular CD therefore. Point O, M and P are collinear.

Clearly, OP=4cm ———-[According to question]

OM=to find?

P is midpoint of AB.

âˆ´AP=1/2 AB=1/2*6=3cm

M is midpoint of AB.

CM=1/2 CD=1/2*8cm=4cm

Join AO and CO

Now in Right âˆ†OPA,

r2=AP2+PO2

r2=32+42

r2=9+16=25

Now in âˆ†OMC

r2=CM2+MO2

25=42+MO2

25-16=MO2

9=MO2

âˆš9=MO

3=MO

âˆ´Therefore distance of the other chord from the centre is 3cm

### Question 4. Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that âˆ ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Solution:

Give: Vertex B of âˆ†ABC lie outside the circle,chord AD=CE

To prove: âˆ ABC=1/2(âˆ DOE-âˆ AOC)

Construction: Join AE

Solution: Chord DE subtends âˆ DOE at the center and âˆ DAE at point A on the circle.

âˆ´âˆ DAE=1/2âˆ DOE       ———-1

chords AC subtends âˆ AOC at the centre and âˆ AEC at point

âˆ´âˆ AEC=1/2âˆ AOC          ———2

In âˆ† ABE,âˆ DAE is exterior angle

âˆ DAE=âˆ ABC +âˆ AEC

1/2âˆ DOE=âˆ ABC+1/2âˆ AOC

Â½(âˆ DOE-âˆ AOC)= âˆ ABC

### Question 5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Solution:

Given: A rhombus ABCD in which O is intersecting point of diagonals AC and BD.

A circle is drawn taking CD as diameter.

To prove: circle points through O or Lies on the circles.

Proof: In rhombus ABCD,

âˆ DOA=90Â°       ——–[diagonals of rhombus intersect at 90Â°] 1

In circle:

âˆ COD=90Â° ——–[angle made in segment  O is right angle] 2

From 1 and 2

O lies on the circle.

### Question 6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

Solution:

ABCD is a ||gm. The  circle through A,B and C intersect at E.

Proof: Here ABCE is a cyclic quadrilateral

âˆ 2+âˆ 4=180Â°      —–[sum of opposite is of a cyclic quadrilateral is 180Â°]

âˆ 4=180Â°-âˆ 1     ——-1

Now âˆ 4+âˆ 6=180Â°-âˆ 6         ———2

From 1 and 2

180Â°-âˆ 2=âˆ 180Â°-âˆ 6

âˆ 2=âˆ 6          ———–3

Also âˆ 2=âˆ 5       ———[opposite angles of ||gm are equal]        —–4

From 3 and 4

âˆ 5=âˆ 6

âˆ 5=âˆ 6

âˆ´AE=AD        ——[sides apposite to equal angles in  aâˆ† are equal]

### Question 7. AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.

Solution:

Given: Two chords AC and BD bisects each other i.e OA=OC,OB=OD

To prove: In âˆ†AOB and COB

AO=CO        ——-[given]

âˆ AOB=âˆ COD    ——[vertically opposite angle]

OB=OD         ——–[give]

âˆ´âˆ†AOBâ‰…âˆ†COB   ——[s.s.s]

AB=CD   ——[C.P.C.T.]         1

similarly âˆ†AODâ‰…âˆ†COB   (S.A.S)

From 1 and 2     ABCD is a ||gm

âˆ´âˆ A+âˆ C=180Â°

âˆ B+âˆ B=180Â°

2âˆ B=180Â°

âˆ B=180Â°/2

âˆ B=90Â°

âˆ´ âˆ A  and âˆ B lies in a semicircle

â†’ AC and BD are diameter of circle.

ii) Since ABCD is a ||gm and âˆ A=90Â°

âˆ´ ABCD is a rectangle.

### Question 8. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90Â° â€“ 1 2 A, 90Â° â€“ 1 2 B and 90Â° â€“ 1 2 C.

Solution:

Given:âˆ†ABC and it circum-circle AD,BE and CF are bisectors of âˆ A,âˆ B and âˆ C

Respectively.

To proof:âˆ  D=90Â°-1/2âˆ A ,    âˆ E=90Â°-1/2âˆ B    ,  âˆ F=90Â°-1/2âˆ C

Construction: Join AE and AF.

Solution: âˆ ADE=âˆ ABE       ———-1  [angle in the same segment are equal]

âˆ ADF=âˆ ACF      ———–2 [angle in the same segment are equal]

âˆ D=1/2âˆ B+1/2âˆ C  ——[BC and CF are bisector of âˆ B & âˆ c]

âˆ D=1/2(âˆ B+âˆ C)

âˆ D=1/2(180Â°-âˆ A)

âˆ D=1/2(180Â°-âˆ A)

âˆ D=90Â°-1/2âˆ AC

### Question 9. Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

Solution:

Given: two congruent circles which intersect at A and B.

PAB is a line segment

To prove: BA=BQ

Construction: join AB

Proof: AB is a common chord of both the congruent circle.

Segment of both circles will be equal

âˆ P=âˆ Q

Now, in âˆ† BPQ,

âˆ P=âˆ Q

BP=BQ       ——[sides opposite to equal angles are equal]

### Question 10. In any triangle ABC, if the angle bisector of âˆ A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Solution:

Given: A âˆ†ABC, in which AD is angle bisector of âˆ A and OD is âŠ¥ bisector of BC.

To prove: D lies on circumcircle.

Construction: Join OB and OC

Proof: Since BC subtends âˆ BAC at A on the remaining of the circle.

âˆ BOC=2âˆ BAC               ——-1

Now, In âˆ†BOE and âˆ† COE

BO=OE      ——–(radii of the same circle)

BE=CE        —–(give)

âˆ´âˆ†BOEâ‰…COE       ——-(S.S.S)

âˆ 1=âˆ 2                     ——-(c.p.c.t)

Now,

âˆ 1+âˆ 2=âˆ BOC

2âˆ 1=âˆ BOC

2âˆ 1=2âˆ BAC         ———- (from 1)

âˆ 1=âˆ BAC

âˆ BOE=âˆ BAF

âˆ BOD=âˆ BAC