# CLASS 9 MATH NCERT SOLUTION FOR CHAPTER – 10 CIRCLES EX – 10.6

## Circles

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 centres O and O’ respectively such that they intersect each other at P and Q.
To Prove: ∠OPO’ = ∠OQO’.
Construction : Join OP, O’P, OQ, O’Q and OO’.
Proof: In ∆OPO’ and ∆OQO’, we have OP = OQ [Radii of the same circle]
O’P = O’Q [Radii of the same circle]
OO’ = OO’ [Common]
∴ AOPO’ = AOQO’ [By SSS congruence criteria]
⇒ ∠OPO’ = ∠OQO’ [C.P.C.T.]

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:
We have a circle with centre O.
AB || CD and the perpendicular distance between AB and CD is 6 cm and AB = 5 cm, CD = 11 cm. Let r cm be the radius of the circle.
Let us draw OP ⊥ AB and OQ ⊥ CD such that
PQ = 6 cm
Join OA and OC.
Let OQ = x cm
∴ OP = (6 – x) cm
∵ The perpendicular drawn from the centre of a circle to chord bisects the chord. 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 ?

Solution:
We have a circle with centre O. Parallel chords AB and CD are such that the smaller chord is 4 cm away from the centre. Let r cm be the radius of the circle and draw OP ⊥ AB and join OA and OC.
∵ OP ⊥ AB
∴ P is the mid-point of AB.
⇒ AP = 12AB = 12(6cm) = 3 cm
Similarly, CQ = 12CD = 12(8cm)= 4 cm
Now in ∆OPA, we have OA2 = OP2 + AP2
⇒ r2 = 42 + 32
⇒ r2 = 16 + 9 = 25
⇒ r = 25−−√ =5
Again, in ∆CQO, we have OC2 = OQ2 + CQ2
⇒ r2 = OQ2 + 42
⇒ OQ2 = r2 – 42
⇒ OQ2 = 52 – 42 = 25 – 16 = 9 [∵ r = 5]
⇒ OQ
⇒ √9 = 3
The distance of the other chord (CD) from the centre is 3 cm.
Note: In case if we take the two parallel chords on either side of the centre, then In ∆POA, OA2 = OP2 + PA2
⇒ r2 = 42 + 32 = 52
⇒ r = 5
In ∆QOC, OC2 = CQ2 + OQ2
⇒ OQ2 = 42 + OQ2
⇒ OQ2 = 52 – 42 = 9
⇒ OQ = 3

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.

Solution:
Given : ∠ABC is such that when we produce arms BA and BC, they make two equal chords AD and CE.
To prove: ∠ABC = 12 [∠DOE – ∠AOC]
Construction: Join AE.
Proof: An exterior angle of a triangle is equal to the sum of interior opposite angles.
∴ In ∆BAE, we have
∠DAE = ∠ABC + ∠AEC ……(i)
The chord DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle. ⇒ ∠ABC = 12 [(Angle subtended by the chord DE at the centre) – (Angle subtended by the chord AC at the centre)]
⇒ ∠ABC = 12 [Difference of the angles subtended by the chords 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.
Taking AB as diameter, a circle is drawn. Let us draw PQ || DA and RS || AB, both are passing through O.
P, Q, R and S are the mid-points of DC, AB, AD and BC respectively,
∵ Q is the mid-point of AB.
⇒ AQ = QB …(i)
Since AD = BC [ ∵ ABCD is a rhombus]
∴ 12 AD = 12 BC
⇒ RA = SB
⇒ RA = OQ …(ii) [ ∵ PQ is drawn parallel to AD and AD = BC]
We have, AB = AD [Sides of rhombus are equal]
⇒ 12 AB = 12 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 of 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.
ABCE is a cyclic quadrilateral.
∴∠AEC + ∠B = 180° …(i)
[Opposite angles of a cyclic quadrilateral are supplementary] But ABCD is a parallelogram. [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 AADE are equal.
∴ Opposite sides must be equal.
⇒ 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:
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
AO = CO [O is the mid-point of AC]
BO = DO [O is the mid-point of BD]
∠AOB = ∠COD [Vertically opposite angles]
∴ Using the SAS criterion of congruence,
∆AOB ≅ ∆COD ⇒ AB = CD [C.P.C.T.]
⇒ arc AB = arc CD …(1)
Similarly, arc AD = arc BC …(2)
Adding (1) and (2), we get
arc AB + arc AD = arc CD + arc BC
⇒ ⇒ BD divides the circle into two equal parts.
∴ BD is a diameter.
Similarly, AC is a diameter.

(ii) We know that ∆AOB ≅ ∆COD
⇒ ∠OAB = ∠OCD [C.P.C.T]
⇒ ∠CAB = ∠ACD
AB || DC
Similarly, AD || BC
∴ ABCD is a parallelogram.
Since, opposite angles of a parallelogram are equal.
∴ ∠DAB = ∠DCB
But ∠DAB + ∠DCB = 180°
[Sum of the opposite angles of a cyclic quadrilateral is 180°]
⇒ ∠DAB = 90° = ∠DCB Thus, ABCD is a rectangle.

Question 8.
Bisectors of angles A, B and C of a ∆ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the ∆DEF are 90° – 12 A, 90° – 12 B and 90° – 12 C.

Solution:
Given : A triangle ABC inscribed in a circle, such that bisectors of ∠A, ∠B and ∠C intersect the circumcise at D, £ and F respectively.
Construction: Join DE, EF and FD.
Proof: ∵ Angles in the same segment are equal.
∴ ∠ED A = ∠FCA …(i)
∠EDA = ∠EBA …(ii)
Adding (i) and (ii), we have
∠FDA + ∠EDA = ∠FCA + ∠EBA
⇒ ∠FDE = ∠FCA + ∠EBA 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:
We have two congruent circles such that they intersect each other at A and B. A line segment passing through A, meets the circles at P and Q.
Let us draw the common chord AB. Since angles subtended by equal chords in the congruent circles are equal.
⇒ ∠APB = ∠AQB
Now, in ∆PBQ, we have ∠AQB = ∠APB
So, their opposite sides must be equal.
⇒ BP = BQ.

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

Solution:
∆ABC with O as centre of its circumcirde. The perpendicular bisector of BC passes through O. Join OB and OC. Suppose it cuts circumcirde at P.
In order to prove that the perpendicular bisector of BC and bisector of angle A of ∆ABC intersect at P, it is sufficient to show that AP is bisector of ∠A of ∆ABC. Arc BC makes angle θ at the circumference
∴ ∠BOC = 2θ
[Angle at centre is double the angle made by an arc at circumference]
Also, in ∆BOC, OB=OC and OP is perpendicular bisector of BC.
So, ∠BOP = ∠COP = θ
Arc CP makes angle θ at O, so it will make
angle [/latex]\frac { \theta }{ 2 }[/latex] at circumference.
So, ∠COP = [/latex]\frac { \theta }{ 2 }[/latex]
Hence, AP is the angle bisedor of ∠A of ∆ABC.

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