Jharkhand Board JAC Class 9 Maths Solutions Chapter 10 Circles Ex 10.6 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.6

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Question 1.

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

Answer:

Given: Two intersecting circles, in which OO’ is the line of centres and P and Q are two points of intersection.

To prove: ∠OPO’ = ∠OQO’

Construction: Join PO, QO, PO’ and QO’.

Proof: In APOO’ and AQOO,’

we have PO = QO [Radii of the same circle]

PO’ = QO'[Radii of the same circle]

OO’ = OO’ [Common]

APOO’ = AQOO’ [SSS axiom]

⇒ ∠OPO’ ≅ ∠OQO’ [CPCT]

Hence, the line of centres of two intersecting circles subtends equal angles at the two points of intersection. Proved.

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.

Answer:

Let O be the centre of the circle and let its radius be r cm.

Draw OM ⊥ AB and OL ⊥ CD.

Then, AM = \(\frac{1}{2}\) AB = \(\frac{5}{2}\) cm

[As perpendicular from centre to the chord bisects the chord]

Similarly, CL = \(\frac{1}{2}\) CD = \(\frac{11}{2}\) cm

Now, LM = 6 cm

Let OL = x cm.

Then OM = (6 – x) cm Join OA and OC.

Then OA = OC = r cm.

Now, from right-angled ∆OMA and ∆OLC, we have

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

and OC^{2} = OL^{2} + CL^{2}

[By Pythagoras Theorem]

r^{2} = (6 – x)^{2} + \(\frac{5}{2}\)^{2}

and r^{2} = x^{2} + \(\frac{11}{2}\)^{2}

⇒ (6 – x)^{2} + \(\frac{5}{2}\)^{2} = x^{2} + \(\frac{11}{2}\)^{2}

⇒ 36 +x^{2} – 12x + \(\frac{25}{4}\) = x^{2} + \(\frac{121}{4}\)

⇒ -12x = \(\frac{121}{4}\) – \(\frac{25}{4}\) – 36

⇒ -12x = \(\frac{96}{4}\) – 36

⇒ -12x = 24 – 36

⇒ -12x = -12

⇒ x = 1

Substituting x = 1 in (i), we get

r^{2} = (6 – x)^{2} + \(\frac{5}{2}\)^{2}

r2 = (6 – 1)^{2} + \(\frac{5}{2}\)^{2}

⇒ r2 = (5)^{2} + \(\frac{5}{2}\)^{2} = 25 + \(\frac{25}{4}\)

⇒ r^{2} = \(\frac{125}{4}\)

⇒ r = \(\frac{5 \sqrt{5}}{2}\) cm

Hence, radius r = \(\frac{5 \sqrt{5}}{2}\) 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 centre?

Answer:

Let PQ and RS be two parallel chord of a circle with centre O.

We have, PQ = 8 cm and RS = 6 cm.

Draw perpendicular bisector OL of RS which meets PQ of M.

Since, PQ || RS, therefore, OM is also perpendicular bisector of PQ.

Also, OL = 4 cm

RL = \(\frac{1}{2}\) RS [As perpendicular from centre to the chord bisects the chord]

= \(\frac{1}{2}\) (6)

= 3 cm

Similarly, PM = \(\frac{1}{2}\) PQ

In ORL, we have

OR^{2} = RL^{2} + OL^{2} [Pythagoras theorem]

⇒ OR^{2} = 3^{2} + 4^{2} = 9 + 16

⇒ OR^{2} = 25

⇒ OP = 5 cm

∴ OR = OP [Radii of the circle]

⇒ OP = 5 cm

Now, in ∆OPM

OM^{2} = OP^{2} – PM^{2} [Pythagoras theorem]

⇒ OM^{2} = 5^{2} – 4^{2} = 25 – 16 = 9

OM = \( \sqrt{9} \) = 3 cm

Hence, the distance of the other chord from the centre is 3 cm.

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

Answer:

Given: Two equal chords AD and CE of a circle with centre O meet at B when produced.

To prove: ∠ABC = \(\frac{1}{2}\) (∠AOC – ∠DOE)

Proof: Let ∠AOC = x, ∠DOE = y, and ∠AOD = ∠EOC = z [Equal chords subtends equal angles at the centre]

∴ x + y + 2z = 360° …(i)

OA = OD ⇒ ∠OAD = ∠ODA

[Angles opposite to equal sides]

∴ In ∆OAD, we have

∠OAD + ∠ODA + z = 180°

⇒ 2 ∠OAD = 180° – z [.’. ∠OAD = ∠ODA]

⇒ ∠OAD = 90° – \(\frac{z}{2}\) …(ii)

Similarly, ∠OCE = 90° – – …(iii)

⇒ ∠ODB = ∠OAD + ∠AOD

[Exterior angle property]

⇒ ∠ODB = 90° – \(\frac{z}{2}\) + z [From (ii)]

⇒ ∠ODB = 90° + \(\frac{z}{2}\) …(iv)

Also, ∠OEB = ∠OCE + ∠COE

[Exterior angle property]

⇒ ∠OEB = 90° – \(\frac{z}{2}\) + z [From (iii)]

⇒ ∠OEB = 90° + \(\frac{z}{2}\) …(v)

In ∆DOE,

OD = OE [Radii of the circle]

⇒ ∠ODE = ∠OED [Angles oposite to equal sides are equal]

Also,

∠ODE + ∠OED + ∠DOE = 180°

[Angle sum property]

⇒ ∠ODE + ∠OED + y = 180°

⇒ 2∠ODE = 180° – y

⇒ ∠ODE = 90° – \(\frac{y}{2}\)

⇒ ∠ODE = ∠OED = 90° – \(\frac{y}{2}\) …(vi)

Now, ∠BED = ∠BEO – ∠OED

= 90 + \(\frac{z}{2}\) – 90 + \(\frac{y}{2}\) [From (v) and (vi)]

= \(\frac{1}{2}\)(y + z)

Also, ∠BDE = ∠BDO – ∠ODE

= 90 + \(\frac{z}{2}\) – 90° + \(\frac{y}{2}\) [From (iv) and (vi)]

= \(\frac{1}{2}\)(y + z)

In ∆BDE,

∠BDE + ∠BED + ∠B = 180°

[Angle sum property]

=> \(\frac{1}{2}\) (y+ z) + \(\frac{1}{2}\) (y + z) + ∠ABC = 180°

⇒ y + z + ∠ABC = 180°

⇒ ∠ABC = 180 – y – z …(vii)

Consider,

\(\frac{1}{2}\)(∠AOC – ∠DOE) = \(\frac{1}{2}\)(x – y)

= \(\frac{1}{2}\)[360° – y – 2z – y] From (i)

= 180° – (y + z) …(viii)

From (vii) and (viii)

∠ABC = \(\frac{1}{2}\) (∠AOC – ∠DOE)

Question 5.

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

Answer:

Given: A rhombus ABCD whose diagonals intersect each other at O.

To prove: A circle with AB as diameter passes through O.

Proof: ∠AOB = 90°

[Diagonals of a rhombus bisect each other at 90°]

⇒ ∆AOB is a right triangle right angled at O.

⇒ AB is the hypotenuse of right ∆AOB.

⇒ If we draw a circle with AB as diameter, then it will pass through O because angle in semicircle is 90° and ∠AOB = 90°.

Question 6.

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

Answer:

Given: ABCD is a parallelogram.

To Prove: AE = AD.

Construction: Draw a circle which passes through ABC and intersect CD produced E.

Proof: As ABCD is a parallelogram,

∠B = ∠ADC …(i)

[Opposite angles of parallelogram]

Also, ∠ADC + ∠ADE = 180°

[Linear Pair]

⇒ ∠B + ∠ADE = 180° …(ii) [From (i)]

Also, ∠B + ∠E = 180° …(iii) [Opposite angles of cyclic quadrilateral]

From (ii), (iii)

∠B = 180° – ∠ADE = 180° – ∠E

⇒ 180° – ∠ADE = 180° – ∠E

⇒ ∠ADE = ∠E

In ∆ADE, AD = AE [Sides opposite to equal angles]

Similarly, we can prove for fig. (ii).

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 rectangle.

Answer:

Given: A circle with chords AC and BD which bisect each other at O.

To Prove: (i) AC and BD are diameters (ii) ABCD is a rectangle.

Proof: In ∆OAB and ∆OCD, we have

OA = OC [Given]

OB = OD [Given]

∠AOB = ∠COD

[Vertically opposite angles]

∠∆AOB = ∆COD [SAS congruence]

∠ABO = ∠CDO [CPCT]

and angles ∠ABO, ∠CDO are alternate interior angles

∴ AB || DC …(i)

Similarly, we can prove BC || AD …(ii)

Hence, ABCD is a parallelogram.

[As opposite sides are parallel]

⇒ ∠A = ∠C [Opposite angles of parallelogram]

and ∠B = ∠D

Also, as ABCD is cyclic

⇒ ∠A + ∠C = 180°

[Opposite angles of cyclic quadrilateral]

⇒ ∠A + ∠A = 180°

⇒ 2∠A = 180°

⇒ ∠A = 90°

So, ABCD is a parallelogram in which one angle is 90°

⇒ ABCD is a rectangle

⇒ ∠ABC = 90° and ∠BCD = 90°.

⇒ AC and BD are diameters.

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° – \(\frac{1}{2}\) A, 90° – \(\frac{1}{2}\) B and 90° – \(\frac{1}{2}\) C.

Answer:

Given: AABC and its circumcircle. AD, BE, CF are bisectors of ∠A, ∠B, ∠C respectively.

Construction: Join DE, EF and FD.

Proof: We know that angles in the same segment are equal.

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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.

Answer:

Given: Two congruent circles which intersect at A and B. PAQ is a line through A.

To Prove: BP = BQ.

Construction: Join AB.

Proof: AB is a common chord of both the circles.

As the circles are congruent,

arc ADB = arc AEB

⇒ ∠APB = ∠AQB [Angles subtended by equal arcs]

So, in APBQ, BP = BQ [Sides opposite to equal angles are equal]

Question 10.

If 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.

Answer:

(i) Let bisector of ∠A meet the circumcircle of ∆ABC at M.

Join BM and CM.

∴ ∠MBC = ∠MAC [Angles in same segment]

and ∠BCM = ∠BAM [Angles in the same segment]

But ∠BAM = ∠CAM [∴ AM is bisector of ∠A]

∴ ∠MBC = ∠BCM So, MB = MC [Sides opposite to equal angles are equal].

⇒ M lies on the perpendicular bisector of BC

Hence, angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of ∆ABC.

(ii) Let M be a point on the perpendicular bisector of BC which lie on circumcircle of ∆ABC

Join AM,

∴ M lies on perpendicular bisector of BC.

∴ BM = CM

∠MBC = ∠MCB [Angle opposite to equal sides are equal]

But ∠MBC = ∠MAC [Angles in the same segment]

and ∠MCB = ∠BAM [Angles in the same segment]

So, from (i)

∠BAM = ∠CAM AM is bisector of ∠A

⇒ bisector of ∠A and perpendicular bisector of BC intersect at M which lies on circumcircle of ∆ABC.