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CHAPTER 10
Circles
Exercise 10.3
1. If arcs AXB and CYD of a circle are congruent, find the ratio of AB and chord CD.
Sol. We have
Since if two arcs of a circle are congruent, then their corresponding arcs are equal, so we
have chord AB = chord CD Hence, AB : CD = 1 : 1.
2. If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at
P and Q, prove that arc PXA ≅ Arc PYB.
Sol. As PQ is the perpendicular bisect of AB,
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So, AM = BM
In and we have
AM = BM [Proved above]
[Each = 90o]
PM = PM [Common side]
[By SAS congruence rule]
So, AP = BP [By C.P.C.T.]
Hence, arc PXA ≅ Arc PYB
[If two chords of a circle are equal, then their corresponding arcs are congruent]
3. A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB,
BC and CA are concurrent.
Sol. Given: Three non-collinear points A, B and C are on a circle.
To prove: Perpendicular bisectors of AB, BC and CA are concurrent.
Construction: Join AB, BC and CA.
Draw perpendicular bisectors ST of AB, PM of BC and QR of CA are respectively. As point A, B
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and C are not collinear, so ST, PM and QR are not parallel and will intersect.
Proof: O lies on ST, the bisector of AB
OA = OB…(1)
Similarly, O lies on PM, the bisector of BC
OB = OC…(2)
And, O lies on QR, the bisector of CA
OC = OA…(3)
From (1), (2) and (3), OA = OB = OC = r(say)
With O as a centre and r as the radius, draw circle C(O, r) which will pass through A, B and C.
This prove that there is a circle passing through the points A, B and C. Since ST, PM or QR can
cut each other at one and only one point O.
O is the only point equidistance from A, B and C.
Hence, the perpendicular bisectors of AB, BC and CA are concurrent.
4. AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC
passes through the centre of the circle.
Sol. Given: AB and AC are two chords which are equal with centre O. AM is the bisector of
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To prove: AM passes through O.
Construction: Join BC. Let AM intersect BC at P.
Proof: In and
AB = AC [Given]
[Given]
And AP = BP [Common side]
[By SAS congruency]
[By C.P.C.T.] And
CP = PB But [Linear
pair ]
AP is perpendicular bisector of the chord BC, which will pass through the centre O on
being produced.
Hence, AM passes through O.
5. If a line segment joining mid-points of two chords of a circle passes through the
centre of the circle, prove that the two chords are parallel.
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Sol. Given: AB and CD are two chords of a circle whose centre of O. The mid-points of AB and
CD are L and M respectively.
To prove: AB||CD
Proof: L is the mid-point of chord AB or
[ The line joining the centre of a circle to the mid-point of a chord is perpendicular to the
chord]
Similarly,
But, these are corresponding angles.
So, AB||CD.
Hence, proved
6. ABCD is such a quadrilateral that A is the centre of the circle passing through B, C
and D. Prove that
Sol. ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D.
We have to prove that
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Join AC.
Since angle subtended by an arc at the centre is double the angle subtended by it at point on
the remaining part of the circle.
Therefore, …(1)
And …(2)
Adding (1) and (2), we get
Hence,
7. O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. Prove
that ∠BOD = ∠A.
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Sol. Given: O is the circumcentre of and
To prove:
Construction: Join OB and OC.
Proof: In and we have
OB = OC [Each equal to radius of the circumcircle]
[Each of 90o]
OD = OD [Common]
[By SAS congruency]
Therefore, BOD = COD [By C.P.C.T.]
Now, are BC subtends at the centre and A at a point in the remaining part of the
circle.
Hence, proved.http
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8. On a common hypotenuse AB, two right triangles ACB and ADB are situated on
opposite sides. Prove that ∠BAC = ∠BDC.
Sol. In right triangles ACB and ADB, we have
and
If the sum of any pair of opposite angles of a quadrilateral is 180o, then the quadrilateral is
cyclic. So, ADBC is a cyclic quadrilateral.
Join CD. Angles and are made by in the same segment BDAC.
Hence,
[ Angles in the same segment of a circle are equal]
9. Two chords AB and AC of a circle subtends angles equal to 90º and 150º, respectively
at the centre. Find ∠BAC, if AB and AC lie on the opposite sides of the centre.
Sol. We have
Reflex
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[Since, angle subtended by an arc at the centre is double of the angle subtended by the same
arc on the remaining part of the cirlce]
Hence, BAC = BOC =
10. If BM and CN are the perpendiculars drawn on the sides AC and AB of the triangle
ABC, prove that the points B, C, M and N are concyclic.
Sol. As BM and CN are the perpendiculars drawn on the sides AC and AB of the triangle ABC
Since, if a line segment (here BC) joining two points (here B and C) subtends equal angles
(here and ) at M and N on the same side of the line (here BC) containing the
segment, the four points (here B, C, M and N) are concyclic.http
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Hence, B, C, M and N are concyclic.
11. If a line is drawn parallel to the base of an isosceles triangle to intersect its equal
sides, prove that the quadrilateral so formed is cyclic.
Sol. is an isosceles triangle in which AB = AC. DE is drawn parallel to BC. We have to
prove that quadrilateral BCED is a cyclic quadrilateral i.e., point B, C, E and D lie on a circle.
In we have
AB = AC[Given]
[ Angles oppossite to equal sides are equal]
Now, DE||BC and AB cuts them,
[ Sum of interior angles on the same side of the transversal]
Similarly, we can show that
Since if pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is
cyclic.
Hence, BCED is a quadrilateral.
12. If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals
are also equal.
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Sol. ABCD is cyclic quadrilateral in which one pair of opposite sides AB = DC. We have to
prove that diagonal AC = diagonal BD.
In and we have
[Angles in the same segment of the circle are equal]
AB = DC[Given]
Also, [Angles in the same segment of the circle are equal]
[By ASA congruence rule]
AO = OD [By C.P.C.T.]…(1)
And OC = BO...(2)
Now, adding (1) and (2),we get
AO + OC = BO + OD
AC = BD
Hence, proved.
13. The circumcentre of the triangle ABC is O. Prove that ∠OBC + ∠BAC = 90º.
Sol. ABC is a triangle and O is the circumcentre.http
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Draw Join OB and OC.
In right and right we have
hyp. OB = hyp. OC
[Radii of the same circle]
OD = OD
[Common side]
[By RHS cong. Rule]
and [By C.P.C.T.]
Now, and
…(1)
[Adding to both sides] [
and ]
Hence, proved.
14. A chord of a circle is equal to its radius. Find the angle subtended by this chord at a
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point in major segment.
Sol. Since chord of a circle is equal radius, so we have AB = OA = OB.
Therefore, ABC is an equilateral triangle.
Since each angle of an equilateral triangle is 60o, so we have Since angle
subtended by an arc at the centre is double the angle subtended by it at any point on the
remaining part of the circle, so we have
Hence,
15. In Fig.10.13, ∠ADC = 130° and chord BC = chord BE. Find ∠CBE.
Sol. In the given figure, we have and chord BC = chord BE. We have to find
Since ABCD is a cyclic quadrilateral and the opposite angles of a cyclic quadrilateral
are supplementary.
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…(1)
In and we have
BC = BE[Given]
OC = OE [Radii of same circle]
OB = OB [Common side]
[By SSS cong. Rule]
[By C.P.C.T. and by (1) ]
Hence,
16. In Fig.10.14, ∠ACB = 40º. Find ∠OAB.
Sol. Since the angle subtended by an arc at the centre is double the angle subtended by it at
any point on the remaining part of the circle, so we have
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So, in we have
OAB = OBA [Angle opposite to equal sides are equal]
Hence,
17. A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC =
130º. Find ∠BAC.
Sol. Since the opposite angles of a cyclic quadrilateral are supplementary.
Now, in [ Angle in a semi-circle = 90o]
And
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18. Two circles with centres O and O′ intersect at two points A and B. A line PQ is drawn
parallel to OO′ through A(or B) intersecting the circles at P and Q. Prove that PQ = 2 OO′.
Sol. Two circles with centre O and O’ intersect at two points A and B. A line PQ is drawn
parallel to OO’ through A (or B) intersecting the circles at P and Q.
Draw and
We have to prove that PQ = 2 OO’.
Since perpendicular from the centre to a chord bisect the chord, so
PA = 2CA…(1)
And AQ = 2 AD…(2)
Adding (1) and (2), we get
PA + AQ = 2CA + 2AD
PQ = 2(CA + AD) = 2CD
Hence, PQ = 2 OO’ [ CD and OO’ are opposite sides of a rectangle]
19. In Fig.10.15, AOB is a diameter of the circle and C, D, E are any three points on the
semi-circle. Find the value of ∠ACD + ∠BED.http
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Sol. Join BC.
Since angle in a semicircle is 90o, we have
∠ACB = 90o
As ABCD is a cyclic quadrilateral and opposite angles of a cyclic quadrilateral are
supplementary
Now, adding to both sides, we get
Hence,
20. In Fig. 10.16, ∠OAB = 30º and ∠OCB = 57º. Find ∠BOC and ∠AOC.ht
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Sol. In we have
OB = OC [Radii of the same circle]
Now, in we have
[ (Given)]
Again, in we have
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