Q.1Â Â Â Â Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Sol.
Given : : Two circles with centres A and B, which intersect each other at C and D.
To prove :
Construction : Join AC, AD, BD and BC.
Proof : In 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,
[C.P.C.T.]
Q.2Â Â Â Â 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.
Sol.
Let O be the centre of the given circle and let its radius be r cm. Draw and . Since , and AB || CD. Therefore, points P, O and Q are collinear. So PQ = 6 cm.
Let OP = x. Then, OQ = (6 â€“ 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 QAP and OCQ, we have
and
............... (1)
and .............. (2)
Â
12x = 60 x = 5
Putting x = 5 in (1), we get
= 5.6 (approx.)
Hence, the radius of the circle is 5.6 cm (approx.)
Q.3Â Â Â Â Â 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.
Sol.
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 and . Since AB || CD and , .
Therefore, points O, Q and P are collinear. Clearly OP = 4 cm, and P, Q are mid-points of AB and CD respectively.
Therefore cm
and, cm
In rt. OAP, we have
r = 5
In rt. OCQ, we have
Â
OQ = 3
Hence, the distance of chord CD from the centre is 3 cm.
Q.4Â Â Â Â Â 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 is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Sol.
Since an exterior angle of a triangle is equal to the sum of the opposite angles.
In BDC, we have
............ (1)
Since angle at the centre is twice the angle at a point on the remaining part of circle.
Therefore
and ................. (2)
From (1) and (2), we have
[Since ]
Hence, is equal to half the difference of angles subtended by the chords AC and DE at the centre.
Q.5Â Â Â Â Â Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Sol.
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
AQ = DP
[Since Q and P are mid-points of AB and CD]
Similarly AE = OQ
AQ = OQ = QB
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.6Â Â Â Â ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Sol.
In order to prove that AE = AD
i.e., AED is an isosceles triangle it is sufficient to prove that . Since ABCE is a cyclic quadrilateral.
Therefore ....... (1)
Now, CDE is a straight line.
............ (2)
[Since and are opposite angles of a parallellogram i.e.
From (1) and (2), we get
Therefore In AED, we have
AD = AE.
Q.7Â Â Â Â Â 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.
Sol.
(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 AOC and BOD, we have
OA = OB [Since O is the mid-point of AB]
[Vert. opp. ]
and, OC = OD [Since O is the mid-point of CD]
By SAS criterion of congruence,
AC = BD
arc(AC) = arc(BD) ............. (1)
Similarly, from AOD and BOC, we have
arc(AD) = arc(BC)
From (1) and (2), we have
arc(AC) + arc(AD) = arc(BD) + arc(BC)
arc(CAD) = arc(CBD)
CD divides the circle into two parts
CD is a diameter.
Similarly, AB is a diameter.
(ii) Since [Proved above]
i.e., i.e.,
AC || BD
Again,
AD || CB
ABCD is a cyclic parallelogram.
.............. (3)
[Opp. of a parallelogram are equal]
Also, ACBD is a cyclic quadrilateral
Therefore ................ (4)
From (3) and (4), we get
Hence, ABCD is a rectangle.
Q.8Â Â Â Â Â 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 , , .
Sol.
We have =
=
=
[Since and are the angles in the same segment of the circle]
Therefore . Similarly, and are the angles in the same segment and hence,
[Since BE is the internal bisector of and CF is the internal bisector
Similarly,
and
and
[Since ]
,
and
The angles of the are , , .
Q.9Â Â Â Â Â 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.
Sol.
Let O and O' be the centre of two congruent circles.
Since AB is a common chord of these circles.
Therefore arc(ACB) = arc(ADB)
BP = BQ.
Q.10 Â Â Â In any triangle ABC, if the angle bisector of 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 and D is the mid-point of BC.
To Prove : DE is the right bisector of BC i.e. .
Construction : Join BE and EC.
Sol.
In and , we have
BE = CE
[Since , therefore arc(BE) = arc(CE) chord BE = chord CE]
BD = CD [Given]
DE = DE [Common]
Therefore by SSS criterion of congruence,
[C.P.C.T.]
Also, [Linear pari]
Therefore
Hence, DE is the right bisector of BC.
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