These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.6

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 of centre O and O’ intersect each other at A and B.

To prove that: ∠OAO’ = ∠OBO’

Construction: Join OO’.

Proof: In ∆AOO’ and ∆BOO’

AO = BO (Radii of the same circle)

AO’ = BO’ (Radii of the same circle)

OO’ = OO’ (Common)

So, by S-S-S congruency condition

∆AOO’ ≅ ∆BOO’

∴ ∠OAO’ = ∠OBO’ (By CPCT)

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 the given circle and let its radius be r cm.

Draw OM ⊥ AB and ON ⊥ CD.

Since OM ⊥ AB, ON ⊥ CD, and AB || CD.

Therefore point M, O and N are collinear. So MN = 6 cm.

Let OM = x cm, Then ON = 6 – x cm

Join OA and OC, then OA = OC = r.

Since the perpendicular from the centre to a chord of the circle bisect the chord.

∴ AM = MB = \(\frac {5}{2}\) cm

and CN = ND = \(\frac {11}{2}\) cm.

In right triangle OAM,

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

r^{2} = x^{2} + \(\left(\frac{5}{2}\right)^{2}\) ……(i)

Now, in right triangle OCN

OC^{2} = ON^{2} + CN^{2}

r^{2} = (6 – x)^{2} + \(\left(\frac{11}{2}\right)^{2}\) …….(ii)

From equation (i) and (ii)

\(x^{2}+\left(\frac{5}{2}\right)^{2}=(6-x)^{2}+\left(\frac{11}{2}\right)^{2}\)

Hence, the radius of the circle is \(\frac{5 \sqrt{5}}{2}\) cm.

Question 3.

The length of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance of 4 cm from the centre, what is the distance of the other chord from the centre.

Solution:

Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 8 cm.

Let the radius of the circle be r cm.

Draw OP ⊥ AB and OQ ⊥ CD.

The length of OP is 4 cm.

In right triangle OAP,

r^{2} = OP^{2} + AP^{2}

⇒ r^{2} = (4)^{2} + (3)^{2} [∵ AP = \(\frac {1}{2}\) AB = 3 cm]

⇒ r^{2} = 16 + 9 = 25

⇒ r = 5 cm ……(i)

Now, in right triangle OQC,

r^{2} = (OQ)^{2} + (CQ)^{2}

⇒ (5)^{2 }= (OQ)^{2} + (4)^{2}

[∵ r = 5 cm prove above and CQ = \(\frac {1}{2}\) CD = 4 cm]

⇒ 25 = (OQ)^{2} + 16

⇒ (OQ)^{2} = 25 – 16 = 9

⇒ OQ = 3 cm

Therefore the distance of the chord CD 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 a circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Solution:

Given: Vertex of an angle ABC be located outside a circle, and AD = CE.

Let ∠AOC = ∠x, ∠AOD = ∠z and ∠DOE = ∠y.

To prove that: ∠ABC = \(\frac {1}{2}\) (x – y)

Proof: AD = CE (given)

∴ ∠AOD = ∠COE (Angle made by equal chord at the centre are equal)

Therefore,

∠x + ∠y + ∠z + ∠z = 360°

⇒ ∠x + ∠y + 2∠z = 360°

⇒ 2∠z = 360° – ∠x – ∠y

⇒ ∠z = 180 – \(\frac {1}{2}\) ∠x – \(\frac {1}{2}\) ∠y ……(A)

Now, ∠ODB = ∠OAD + ∠DOA

(Exterior angle is equal to sum of opposite interior angles)

∠ODB = ∠OAD + ∠z ……(i)

Again, ∠OAD + ∠z + ∠ODA = 180°

(Sum of all three angles of triangles)

or, ∠OAD + ∠z + ∠OAD = 180°

(Angle opposite to equal sides are equal and OA = OD radii of circle)

or, 2∠OAD = 180° – ∠z

or, ∠AOD = \(\frac {1}{2}\) (180° – ∠z) = 90° – \(\frac {1}{2}\) ∠z

Putting the value of ∠AOD in equation (i)

∴ ∠ODB = 90° – \(\frac {1}{2}\) ∠z + ∠z

(∵ ∠OAD = 90° – \(\frac {1}{2}\) ∠z)

or, ∠ODB = 90° + \(\frac {1}{2}\) ∠z ……..(ii)

Similarly, ∠OEB = 90° + \(\frac {1}{2}\) ∠z ……(iii)

Now, in ODBE

∠ODB + ∠B + ∠OEB + ∠y = 360°

(Sum of all angle of ∆ is 360°)

90 + \(\frac {1}{2}\) ∠z + ∠B + 90 + \(\frac {1}{2}\) ∠z + ∠y = 360°

(Use equation (ii) and (iii))

or, 180 + ∠z + ∠B + ∠y = 360

or, ∠B = 180 – ∠y – ∠z …..(iv)

⇒ ∠ABC = \(\frac {1}{2}\) [(Angle subtended by the chorde DE at the centre) – (Angle subtended by the chord AC at the centre)]

⇒ ∠ABC = \(\frac {1}{2}\) [(Difference of the angles subtended by the chorde DE and AC at the centre)]

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:

We have a rhombus ABCD such that its diagonals AC and BD intersect at O.

P, Q, R and S are the mid-points of DC, AB, AD and BC respectively

∴ \(\frac {1}{2}\) AD = \(\frac {1}{2}\) BC

⇒ RA = SB

⇒ RA = OQ ……(ii)

[∴ PQ is drawn parallel to AD and AD = BC]

⇒ \(\frac {1}{2}\) AB = \(\frac {1}{2}\) AD

⇒ AQ = AR ……(iii)

From (i), (ii) and (iii), we have

AQ = QB = OQ

i.e., A circle drawn with Q as centre, will pass through A, B and O.

Thus, the circle passes through the intersection ‘O’ of the diagonals rhombus ABCD.

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:

We have a circle passing through A, B, and C is drawn such that it intersects CD at E.

∴ ∠AEC + ∠B = 180° ……(i)

[Opposite angles of a cyclic quadrilateral are supplementary]

But ABCD is a prallelogram (Given)

∴ ∠D = ∠B ……(ii)

[Opposite angles of a parallelogram are equal]

From (i) and (ii), we have

∠AEC + ∠D = 180° …….(iii)

But ∠AEC + ∠AED = 180° (Linear pair) ……(iv)

From (iii) and (iv),

We have ∠D = ∠AED

i.e., The base angles of ∆ADE are euqal.

∴ AE = AD

Question 7.

AC and BD are chords of a circle which bisects each other. Prove that

(i) AC and BD are diameters

(ii) ABCD is a rectangle.

Solution:

Given: A circle in which two chords AC and BD are such that they bisect each other.

Let their point of intersection be O.

To Prove: (i) AC and BD are diameters.

(ii) ABCD is a rectangle.

Construction: Join AB, BC, CD, and DA.

Proof: (i) In ∆AOB and ∆COD, we have

(∵ ∠ADC = ∠ABC, opposite angles of || gm ABCD)

From equation (i) and (ii)

∠AED + ∠ABC = ∠ADE + ∠ABC

or, ∠AED = ∠ADE

Thus, in ∆AED, we have

∠AED = ∠ADE

∴ AD = AE. (Side opposite to equal angles are equal)

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 angle of the triangle DEF are 90° – \(\frac {1}{2}\) ∠A, 90° – \(\frac {1}{2}\) ∠B and 90° – \(\frac {1}{2}\) ∠C.

Solution:

Given: In ∆ABC, AD, BE and CF is the angle bisector of ∠A, ∠B, and ∠C respectively.

Where D, E, and F lie on the circumcircle of ∆ABC.

To prove that:

(i) ∠FDE = 90° – \(\frac {1}{2}\) ∠A

(ii) ∠DEF = 90° – \(\frac {1}{2}\) ∠B

(iii) ∠EFD = 90° – \(\frac {1}{2}\) ∠C

Proof:

(i) ∠ADF = ∠ACF ……(i) (Angles in the same segments of a circle are equal)

Again, ∠ADE = ∠ABE ……(ii) (Angles in the same segment of atcircle are equal)

Adding (i) and (ii)

∠APF + ∠ACF = ∠ACF + ∠ABE

or, ∠FDE = \(\frac {1}{2}\) ∠C + \(\frac {1}{2}\) ∠B ……(iii)

(∵ CF and BE are the bisector of ∠B and ∠C respectively)

Now, In ∆ABC

∠A + ∠B + ∠C = 180° (Sum of all angles of a ∆ is 180°)

or, \(\frac {1}{2}\) (∠A + ∠B + ∠C) = 90°

or, \(\frac {1}{2}\) ∠B + \(\frac {1}{2}\) ∠C = 90 – \(\frac {1}{2}\) ∠A ……(iv)

From equation (iii) and (iv)

∠FDE = 90° – \(\frac {1}{2}\) ∠A

Similarly we can prove

(ii) ∠DEF = 90 – \(\frac {1}{2}\) ∠B

(iii) ∠EFD = 90 – \(\frac {1}{2}\) ∠C.

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:

Let C(0, r) and C(O’, r) be two congruent circles.

Since AB is a common chord of two congruent circles. Therefore,

arc ACB = arc ADB

∴ ∠BPA = ∠BQA

Thus, in ∆BPQ, we have ∠BPA = ∠BQA

∴ BP = BC (Sides opposite to equal angles of a triangle are equal)

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:

Given: ABC is a triangle in which the bisector of ∠A intersects the circumcircle of ∆ABC at D.

To prove that: Perpendicular bisector of side BC intersects the angle bisector of ∠A at D.

Construction: Join BD and CD.

Proof: ∠BCD = ∠BAD ……(i)

(Angles in the same segment of a circle are equal)

Again, ∠DBC = ∠DAC …….(ii)

(Angles in the same segment of a circle are equal)

But, ∠BAD = ∠DAC …..(iii)

(∵ AD is the bisector of ∠A)

From equation (i), (ii) and (iii)

∠BCD = ∠DBC

or, BD = CD (Side opposite to equal angles of a ∆ are equal)

So, D must be lying on the perpendicular bisector of BC.

Therefore, the perpendicular bisector of side BC intersects the angle bisector of ∠A).