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**Karnataka State Syllabus Class 9 Maths Chapter 12 Circles Ex 12.6**

## Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.6

Question 1.

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

Solution:

Data: Two circles having centres A and B, intersect at C and D.

To Prove: ∠ACB = ∠ADB.

Construction: Join A and B.

Proof: In ∆ABC and ∆ABD,

AC = AD (∵ radii of same circle are equal)

BC = DD

AB is common.

∴ ∆ABC ≅ ∆ABD (SSS Postulate.)

∴ ∠ACB = ∠ADB.

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:

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.

Distance between AB and CD is 6 cm.

To Prove: Radius of the circle, OP =?

Construction: Join OP and OQ, OB, and OD.

Proof: Chord AB || Chord CD.

AB = 5 cm, and CD =11 cm.

OP ⊥ AB

∴ BP = AP = 52 = 2.5 cm.

OQ⊥CD

∴ CQ = QD = 112 = 5.5 cm.

PQ = 6 cm. (Data)

Let OQ = 2 cm then, OP = (6 – x) cm.

In ∆BPO, ∠P = 90°

As per Pythagoras theorem,

OB^{2} = BP^{2} + PO^{2}

= (2.5)^{2} + (6 – x)^{2}

= 6.25 + 36 – 12x + x^{2}

OB^{2} = x^{2}– 12x + 42.25 …………….. (i)

In ∆OQD, ∠Q= 90°

∴ OD^{2} = OQ^{2} + QD^{2}

= (x)^{2} + (5.5)^{2}

OD^{2} = x^{2} + 30.25 ……………….. (ii)

OB = OD (∵ radii of same circle)

From (i) and (ii).

x^{2} – 12x + 42.25 = x^{2} + 30.25

-12x = 30.25 – 42.25

-12x = -12

12x = 12

∴ x = 1212

∴ x = 1 cm.

From (ii),

OD^{2} = x^{2} + 30.25

= (1)^{2} + 30.25

= 1 + 30.25

∴ OD^{2} = 31.25

OD = 31.25−−−−√

∴ OD = 5.59 cm.

∴ Radius of circle OP = OD = 5.59 cm.

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 centre, what is the distance of the other chord from the center?

Solution:

Data: Chords of a circle are 6 cm. and 8 cm. are parallel. The smaller chord is at distance of 4 cm. from the centre.

To Prove: Distance between bigger chord and centre =?

Construction: Join OA and OC.

Proof: AB || CD, AB = 6 cm, CD = 8 cm.

OP⊥CD, OQ⊥CD.

In ∆OPA, ∠P = 90°

∴ OA^{2} = OP^{2} + PA^{2} (According to Pythagoras theorem)

= (4)^{2 }+ (3)^{2} = 16 + 9

OA^{2} = 25

∴ OA = 5 cm.

OA = OC = 5 cm. (radii of the same circle.)

Now, in ∆OQC,

OC^{2} = OQ^{2} + QC^{2}

(5)^{2} = x^{2} + (4)^{2}

25 = x^{2} + 16

x^{2} = 25 – 16 = 9

∴ x = 9–√ ∴ x = 3 cm.

∴ Bigger chord is at a distance of 3 cm from the centre.

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 ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Solution:

Data: The vertex of angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle.

To Prove: ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre. OR

∠ABC= 12 [∠DOE – ∠AOC].

Construction: OA, OC, OE, OD are joined.

Proof: In ∆AOD and ∆COE,

OA = OC, OD = OE radii of same circle.

AD = CE (Data)

∴ ∆AOD ≅ ∆COE (SSS Postulate)

∠OAD = ∠OCE ……….. (i)

∴∠ODA = ∠OEC …………. (ii)

OA = OD

∴ ∠OAD = ∠ODA …………. (iii)

From (i) and (ii),

∠OAD = ∠OCE = ∠ODA = ∠OEC = x°.

In ∆ODE, OD = OE

∠ODE = ∠OED = y°.

ADEC is a cyclic quadrilateral.

∴ ∠CAD + ∠DEC = 180°

x + a + x + y = 180

2x + a + y = 180

y = 180 – 2x – a ……….. (iv)

But, ∠DOE = 180 – 2y

∠AOC = 180 – 2a

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:

Data: ABCD is a rhombus. Circle is drawn taking side CD diameter. Let the diagonals AC and BD intersect at ‘O’.

To Prove: Circle passes through the point ‘O’ of the intersection of its diagonals.

Proof: ∠DOC = 90° (Angle in the semicircle) and diagonals of rhombus bisect at right angles at ‘O’.

∴ ∠DOC = ∠COB = ∠BOA = ∠AOD = 90°

∴ Circle passes the point of intersection of its diagonal through ‘O’.

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:

Data: ABCD is a parallelogram. The circle through A, B, and C intersect CD at E. AE is joined.

To Prove: AE = AD

Proof: ∠AEC + ∠AED = 180° …………. (i) (linear pair)

ABCE is a cyclic quadrilateral.

∴ ∠ABC + ∠AEC = 180° ………….(ii) (opposite angles)

Comparing (i) and (ii),

∠AEC = ∠AED = ∠ABC + ∠AEC

∠AED = ∠ABC ………….. (iii)

But, ∠ABC = ∠ADE (Opposite angles of quadrilateral)

Substituting in equation (iii),

∠AED = ∠ADE

∴ AE = AD.

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:

Data : AC and BD are chords of a circle bisect each other at ‘O’.

To Prove:

i) AC and BD are diameters.

ii) ABCD is a rectangle.

Construction: AB, BC, CD and DA are joined.

Proof: In ∆AOB and ∆COD,

AO = OC (Data)

BO = OD (Data)

∠AOB = ∠COD (vertically opposite angles)

∴ ∆AOB ≅ ∆COD (SAS postulate)

∴ ∠OAB = ∠OCD

These are pair of alternate angles.

∴ AB || CD and AB = CD.

∴ ABCD is a parallelogram.

∴ ∠BAD = ∠BCD (Opposite angles of parallelogram)

But. ∠BAD + ∠BCD = 180 (∵ Angles of cyclic quadrilateral)

∠BAD + ∠BAD = 180

2(∠BAD) = 180

∴ ∠BAD = 1802

∴ ∠BAD = 90°.

If angles of a quadrilateral are right angles it is rectangle. ABCD is a recrtangle.

∠BAD = 90°

∠BAD is separated from chord BD.

∴ This is the angl in semicircle.

∴ Chord BD is a diameter.

Similarly, ∠ADC = 90°

∴ Chord AC is a diameter.

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 90° – 12 A, 90° – 12 B and 90° – 12 C.

Solution:

Data: AD, BE and CF are angular bisectors of angles A, B and C of ∆ABC intersects its circumference at D, E and F respectively.

To prove: Angles of ∆DEF are 90° – 12 A, 90° – 12 B and 90° – 12 C.

Proof: AD, BE and CF are angular bisectors of angles A, B and C of ∆ABC.

∴ ∠BAD = ∠CAD = ∠A2

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 lies on the two circles.

Prove that BP = BQ.

Solution:

Data : 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.

To Prove: BP = BQ

Construction: Join AB.

Proof: Two congruent triangles with centres O and O’ intersects at A and B. Through A segment PAQ is drawn so that P, Q lie on the two circles.

Similarly, ∠AQB= 70° in circle subtended by chord AB. Because Angles subtended by circumference by same chord.

∴ ∠APB = ∠AQB = 70°.

Now, in ∆PBQ, ∠QPB = ∠PQB.

∴ Sides opposite to each other are equal.

∴ BP = BQ.

Question 10.

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

Solution:

Data: In ∆ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect each other. O is the centre of the circle.

To Prove: Angle bisector of ∠A and perpendicular bisector of BC intersect at D.

Construction: Join OB, OC.

Proof: Angle subtended at the Centre

= 2 × angles subtended in the circumference.

∠BOC = 2 × ∠BAC

In ∆BOE and ∆COE,

∠OEB = ∠OEC = 90° (∵ OE⊥BC)

∴ BO = OC (radii)

OE is common.

∴ ∆BOE ≅ ∆COE (RHS postulate)

But, ∠BOE + ∠COE = ∠BOC

∠BOE + ∠BOE = ∠BOC

2∠BOE = ∠BOC

2∠BOE = 2∠BAC

∴ ∠BOE = ∠BAC

But, ∠BOE = ∠COE = ∠BAC

∠BAD = 12 ∠BAC

∠BAD = 12 ∠BOE

∠BAD = 12 ∠BOD

∴ ∠BOD = 2∠BAD

∴ The angle subtended by an arc at the centre is double the angle subtended by it at any point on the circumference.

∴ Angle bisector of ∠A and perpendicular bisector of BC intersect at D.

**All Chapter KSEEB Solutions For Class 9 Maths**

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