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Circles : Exercise 10.6 (Optional) (Mathematics NCERT Class 9th)

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

Given : : Two circles with centres A and B, which intersect each other at C and D.

To prove : \angle ACB = \angle ADB
Construction : Join AC, AD, BD and BC.
Proof : In \Delta s ACB and ADB, we have
AC = AD [Radii of the same circle]
BC = BD [Radii of the same circle]
AB = AB [Common]
Therefore by SSS criterion of congruence,
\Delta ACB \cong \Delta ADB
 \Rightarrow \angle ACB = \angle ADB [C.P.C.T.]

Q.2     Two chords AB and CD of lengths 5 cm and 11 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.

Let O be the centre of the given circle and let its radius be r cm. Draw OP \bot AB and OQ \bot CD. Since OP \bot AB, OQ \bot CD and AB || CD. Therefore, points P, O and Q are collinear. So PQ = 6 cm.
Let OP = x. Then, OQ = (6 – x) cm.
Join OA and OC. Then, OA = OC = r.
Since the perpendicular from the centre to a chord of the circle bisects the chord.
Therefore AP = PB = 2.5 cm and CQ = QD = 5.5 cm.
In right \Delta s QAP and OCQ, we have
O{A^2} = O{P^2} + A{P^2} and O{C^2} = O{Q^2} + C{Q^2}
 \Rightarrow {r^2} = {x^2} + {\left( {2.5} \right)^2} ............... (1)
and {r^2} = {\left( {6 - x} \right)^2} + {\left( {5.5} \right)^2} .............. (2)
 \Rightarrow {x^2} + {\left( {2.5} \right)^2} = {\left( {6 - x} \right)^2} + {\left( {5.5} \right)^2}
 \Rightarrow  {x^2} + 6.25 = 36 - 12x + {x^2} + 30.25

 \Rightarrow 12x = 60  \Rightarrow x = 5
Putting x = 5 in (1), we get
{r^2} = {5^2} + {\left( {2.5} \right)^2} = 25 + 6.25 = 31.25
 \Rightarrow r = \sqrt {31.25} = 5.6 (approx.)
Hence, the radius of the circle is 5.6 cm (approx.)

Q.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.

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 \bot AB and OQ \bot CD. Since AB || CD and OP \bot AB, OQ \bot CD.
Therefore, points O, Q and P are collinear. Clearly OP = 4 cm, and P, Q are mid-points of AB and CD respectively.
Therefore AP = PB = {1 \over 2}AB = 3 cm
and, CQ = QD = {1 \over 2}CD = 4 cm
In rt. \angle d \Delta OAP, we have
O{A^2} = O{P^2} + A{P^2}
 \Rightarrow {r^2} = {4^2} + {3^2} = 16 + 9 = 25
 \Rightarrow r = 5
In rt. \angle d \Delta OCQ, we have
O{C^2} = O{Q^2} + C{Q^2}
 \Rightarrow {r^2} = O{Q^2} + {4^2}
 \Rightarrow  25 = O{Q^2} + 16
 \Rightarrow O{Q^2} = 9
 \Rightarrow OQ = 3

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

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


Since an exterior angle of a triangle is equal to the sum of the opposite angles.
In \Delta BDC, we have
\angle ADC = \angle DBC + \angle DCB ............ (1)
Since angle at the centre is twice the angle at a point on the remaining part of circle.
Therefore \angle ADC = {1 \over 2}\angle AOC
and \angle DCB = {1 \over 2}\angle DOE ................. (2)
From (1) and (2), we have
{1 \over 2}\angle AOC = \angle ABC + {1 \over 2}\angle DOE [Since \angle DBC = \angle ABC]
 \Rightarrow \angle ABC = {1 \over 2}\left( {\angle AOC - \angle DOE} \right)
Hence, \angle ABC is equal to half the difference of angles subtended by the chords AC and DE at the centre.

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

Given : ABCD is a rhombus. AC and BD are its two diagonals which bisect each other at right angles.
To prove : A circle drawn on AB as diameter will pass through O.
Construction : From O draw PQ || AD and EF || AB.
Proof : AB = DC  \Rightarrow {1 \over 2}AB = {1 \over 2}DC
[Since Q and P are mid-points of AB and CD]
Similarly AE = OQ
 \Rightarrow AQ = OQ = QB
 \Rightarrow A circle drawn with Q as centre and radius AQ passes through A, O and B.
The circle thus obtained is the required circle.

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

In order to prove that AE = AD
i.e., \Delta AED is an isosceles triangle it is sufficient to prove that \angle AED = \angle ADE. Since ABCE is a cyclic quadrilateral.
Therefore \angle AED + \angle ABC = 180^\circ ....... (1)
Now, CDE is a straight line.
 \Rightarrow \angle ADE + \angle ADC = 180^\circ ............ (2)
[Since \angle ADC and \angle ABC are opposite angles of a parallellogram i.e. \angle ADC = \angle ABC
From (1) and (2), we get
\angle AED + \angle ABC = \angle ADE + \angle ABC
 \Rightarrow \angle AED = \angle ADE
Therefore In \Delta AED, we have
\angle AED = \angle ADE
 \Rightarrow AD = AE.

Q.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.

(i) Let AB and CD be two chords of a circle with center O.
Let they bisect each other at O.
Join AC, BD, AD and BC.
In \Delta s AOC and BOD, we have
OA = OB [Since O is the mid-point of AB]
\angle AOC = \angle BOD [Vert. opp. \angle s]
and, OC = OD [Since O is the mid-point of CD]
By SAS criterion of congruence,
\Delta AOC \cong \Delta BOD
 \Rightarrow AC = BD
 \Rightarrow arc(AC) = arc(BD) ............. (1)
Similarly, from \Delta s AOD and BOC, we have
arc(AD) = arc(BC)
From (1) and (2), we have
arc(AC) + arc(AD) = arc(BD) + arc(BC)
 \Rightarrow arc(CAD) = arc(CBD)
 \Rightarrow CD divides the circle into two parts
 \Rightarrow CD is a diameter.
Similarly, AB is a diameter.
(ii) Since \Delta AOC \cong \Delta BOD [Proved above]
 \Rightarrow \angle OAC i.e., \angle BAC = \angle OBD i.e., \angle ABD
 \Rightarrow AC || BD
Again, \Delta AOD \cong \Delta COB
 \Rightarrow AD || CB
 \Rightarrow ABCD is a cyclic parallelogram.
 \Rightarrow \angle DAC = \angle DBA .............. (3)
[Opp. \angle s of a parallelogram are equal]
Also, ACBD is a cyclic quadrilateral
Therefore \angle DAC + \angle DBA = 180^\circ ................ (4)
From (3) and (4), we get
\angle DAC = \angle DBA = 90^\circ
Hence, ABCD is a rectangle.

Q.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}A, 90^\circ - {1 \over 2}B, 90^\circ - {1 \over 2}C.

We have \angle D = \angle EDF
= \angle EDA + \angle ADF
= \angle EBA + \angle FCA
[Since \angle EDA and \angle EBA are the angles in the same segment of the circle]
Therefore \angle EDA = \angle EBA. Similarly, \angle ADF and \angle FCA are the angles in the same segment and hence, \angle ADF = \angle FCA
 = {1 \over 2}\angle B + {1 \over 2}\angle C
[Since BE is the internal bisector of \angle B and CF is the internal bisector \angle C
\angle D = {{\angle B + \angle C} \over 2}
Similarly, \angle E = {{\angle C + \angle A} \over 2}
and \angle F = {{\angle A + \angle B} \over 2}
 \Rightarrow  \Rightarrow \angle D = {{180^\circ - \angle A} \over 2}
\angle E = {{180^\circ - \angle B} \over 2}
and \angle F = {{180^\circ - \angle C} \over 2}
[Since \angle A + \angle B = 180^\circ ]
 \Rightarrow \angle D = 90^\circ - {{\angle A} \over 2},
\angle E = 90^\circ - {{\angle B} \over 2}
and \angle F = 90^\circ - {{\angle C} \over 2}
 \Rightarrow The angles of the \Delta DEF are 90^\circ - {1 \over 2}A, 90^\circ - {1 \over 2}B, 90^\circ - {1 \over 2}C.

Q.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.

9Let O and O' be the centre of two congruent circles.
Since AB is a common chord of these circles.
Therefore arc(ACB) = arc(ADB)
\angle BPA = \angle BQA

 \Rightarrow BP = BQ.

Q.10     In any triangle ABC, if the angle bisector of \angle A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Given : ABC is a triangle inscribed in a circle with centre at O, E is a point on the circle such that AE is the internal bisector of \angle BAC and D is the mid-point of BC.

To Prove : DE is the right bisector of BC i.e. \angle BDE = \angle CDE = 90^\circ .
Construction : Join BE and EC.

In \Delta BDE and \Delta CDE, we have
[Since \angle BAE = \angle CAE, therefore arc(BE) = arc(CE)  \Rightarrow chord BE = chord CE]
BD = CD [Given]
DE = DE [Common]
Therefore by SSS criterion of congruence,
\Delta BDE \cong \Delta CDE
 \Rightarrow \angle BDE = \angle CDE [C.P.C.T.]
Also, \angle BDE + \angle CDE = 180^\circ [Linear pari]
Therefore \angle BDE = \angle CDE = 90^\circ
Hence, DE is the right bisector of BC.


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