Circles – Exercise 10.6 – (MATHEMATICS) – 9th Class

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Exercise 10.6

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


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Sol. Consider triangles APB and AQB,

AP = AQ                               [Radii]

PB = QB                               [Radii]

AB is common

∴ ΔAPB ≅  ΔAQB                                  [SSS]

∴   ∠APB = ∠AQB.                                [CPCT]


 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.

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Sol.  Let OL is drawn perpendicular to AB and LO is produced to M meeting CD at M, then LM ⊥ CD.
[ AB ǁ‖ CD]

∴   ∠ALM = 90° ; ∠OMC = 90°

AL = {1 \over 2}AB = {5 \over 2}cm;

CM = {1 \over 2}CD = {{11} \over 2}cm.

Let OL = x, OA = r. Then OM = 6 – x.

Now, in right-angled triangle OLA.

O{A^2} = A{L^2} + O{L^2} = {\left( {{5 \over 2}} \right)^2} + {X^2}

 \Rightarrow {r^2} = {{25} \over 4} + {x^2}                                 ……(i)

In right-angled triangle OML,

O{C^2} = C{M^2} + O{M^2} = {\left( {{{11} \over 2}} \right)^2} + {(6 - x)^2}

 \Rightarrow {r^2} = {{121} \over 4} + 36 - 12x + {x^2}                   ……(ii)

From (i) and (ii), we get

{{25} \over 4} + {x^2} = {{121} \over 4} + 36 - 12x + {x^2}

 \Rightarrow 12x = {{121} \over 4} + 36 - {{25} \over 4} = 6 \Rightarrow x = 5

Substituting x = 5 in (i), we get

{r^2} = {{25} \over 4} + {(5)^2} = {{25} \over 4} + 25 = {{125} \over 4}

 \Rightarrow r = {{5\sqrt 5 } \over 2}cm

Hence, radius of the circle is {{5\sqrt 5 } \over 2}cm .


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?

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Sol. Let AB = 6 cm and CD = 8 cm are two parallel chords of a circle with centre at O.
We draw OM perpendicular to AB meeting AB at M and CD at L. Then OL is also perpendicular to CD.

AM = {1 \over 2}AB = {1 \over 2} \times 6 = 3

and CL = {1 \over 2}CD = {1 \over 2} \times 8 = 4cm.

Let OL = x.

In right-angled triangle OMA,

OA2 = AM² + OM² = (3)² + (4)² = 25    ⇒  OA = 5 cm.

As OC = OA  ∴ OC = 5 cm.

In right-angled triangle OLC,

OC² = CL² + OL²  ∴ (5)² = (4)² + x²

⇒  x² = 25 – 16 = 9    ∴    x = 3

Therefore, distance of other chord is 3 cm.


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.

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Sol. Construction: Join E and A.
Proof: ∠AOC = 2∠AEC                 …(i)

[Angle subtended by an arc at the centre is double the angle subtended by it on the remaining part of the circle.]

∠DOE = 2∠DAE                                     …(ii)                              [Reason same as above]

Also ∠AEC = ∠ABC + ∠BAE

[Exterior angle of a triangle is equal to sum of interior opposite angles]

⇒  ∠AEC = ∠ABC + ∠DAE

 \Rightarrow {1 \over 2}\angle AOC = \angle ABC + {1 \over 2}\angle DOE

⇒  \angle ABC = {1 \over 2}(\angle AOC - \angle DOE)


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

 

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Sol. We know that diagonals of a rhombus intersect each other at right angles.

∴  ∠AOB = 90°. Also we know that angle in a semicircle is 90°. If a circle is drawn with AB as diameter, then O lies on the circle.

Hence, if a circle is drawn with any sides of the rhombus as diameter, then it passes through the point of intersection of diagonals.


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

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Sol. ABCD is a parallelogram.
∴  ∠ABC =  ∠CDA …(i)                                                        [Opposite angles of a parallelogram]

Also, ABCE is a cyclic quadrilateral.

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

[Sum of opposite angles of a cyclic quadrilateral is 180°]

In Fig. I,
Also, ∠ADC + ∠ADE = 180°                                                    [Linear pair]

⇒     ∠ABC + ∠ADE = 180° .                                                  ..(iii) [From (i)]

From (ii) and (iii), we get

∠AEC = ∠ADE    ⇒   ∠AED = ∠ADE

⇒     AD = AE
[Sides opposite to equal angles of a triangle are equal.]

 

In Fig. II,
∠AEC + ∠AED = 180° …(iv)                                                [Linear pair]

From (i) and (ii), we get

∠CDA + ∠AEC = 180°    ⇒    ∠EDA + ∠AEC = 180°                                      …(v)

From (iv) and (v), we get

∠AED = ∠EDA

⇒               AD = AE.
[Sides opposite to equal angles of a triangle are equal]


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.

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Sol. (i) Consider triangles AOD and BOC.

AO = OC [Given]

BO = OD [Given]

∠AOD = ∠BOC                          [Vertically opposite angles]

⇒   ΔAOD   ≅   ΔCOB                            [SAS]

⇒  ∠DAO = ∠BCO

Also,  ∠ADO = ∠BCO

[Angles in the same segment of a circle]
From (i) and (ii), we have

∠DAO = ∠ADO

⇒  OA = OD

Hence, OA = OB = OC = OD.

⇒   0 is equidistant from A, B, C, D.

⇒   0 is centre of the circle.

Hence, AC and BD are diameters.

(ii) As AC and BD are diameters. Therefore, ∠BAD,∠ABC, ∠BCD, ∠CDA are angles in a semicircle.
Therefore, each of them is 90°.
Hence, ABCD is a rectangle.


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^\circ - {1 \over 2}\angle A,90^\circ - {1 \over 2}\angle B  and

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Sol. ∠ADE = ∠ABE                                       …(i)
[Angles in the same segment of a circle]
∠ADF = ∠ACF                                                …(ii) [Reason same as above]

Now ∠EDF = ∠ADE + ∠ADF
⇒  ∠EDF = ∠ABE + ∠ACF

…(iii) [From (i), (ii)]

As BE and CF are bisectors of ∠ABC and ∠ACB.

\angle ABE = {1 \over 2}\angle ABC  and \angle ACF = {1 \over 2}\angle ACB   ….(iv)

From (iii) and (iv), we get

\angle EDF = {1 \over 2}\angle ABC + {1 \over 2}\angle ACB = {1 \over 2}(\angle ABC + \angle ACB)                                           …..(v)

In triangle ABC,

∠ABC + ∠ACB + ∠BAC   =180°

⇒ ∠ABC + ∠ACB  = 180° – ∠BAC.

Substituting in (v), we get

\angle EDF = {1 \over 2}(180^\circ - \angle BAC) = 90^\circ - {1 \over 2}\angle BAC = 90^\circ - {1 \over 2}\angle A

Similarly, we can show that
\angle DEF = 90^\circ - {1 \over 2}\angle B;\angle DFE = 90^\circ - {1 \over 2}\angle C


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.

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Sol. As AB is common chord.


∴   arc AXB = arc AYB
⇒  ∠AQB = ∠APB
[Equal areas of congruent circles subtend equal angles on the remaining part of the circles.]
Now in triangle BPQ,
∠AQB = ∠APB                                                [Proved above]
⇒  ∠PQB = ∠QPB
⇒  PB = QB.

[Sides opposite to equal angles are equal.]


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

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Sol. Let 0 be the centre of the circumcircle of triangle ABC.
∴   ∠BOC = 2∠BAC
[Angle subtended by an arc at the centre is doubled the angle subtended by it at remaining part of the circle.]

Consider ΔOLB and ΔOLC.

OB = OC                                                                     [Radii]

OC is common.

BL = LC                                                                  [OL is perpendicular bisector]

∴   ΔOLB  ≅  ΔOLC                                               [SSS]

∴  ∠BOL = ∠COL = ∠CAB                                       …(i)

Let perpendicular bisector of side BC and angle bisector ∠A meet a the point P.
From equation (i), we get

∠COL = ∠CAB

⇒   ∠COP = 2∠CAP                                                            [∠CAP = ∠BAP]

This proves that points C, A and P are concyclic points as ∠COP is subtended by arc CP at the centre O and ∠CAP is subtended by it at the remaining part of the circle.
Again, from equation (i), we get

∠BOL = ∠BAC

i.e., ∠BOP = 2∠BAP

This, also proves that points B, A and P are concyclic. Hence, P lies on the circumcircle of the ΔABC.


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