1

ASSIGNMENT

LEVEL - I
1. In the given figure, three circles with centres A, B, C respectively touch each other externally. If AB
= 5 cm, BC = 7 cm and CA = 6 cm, then the radius of the circle with centre A is

(a) 1.5 cm (b) 2 cm (c) 2.5 cm (d) 3 cm

2. In Fig., if AD, AE and BC are tangents to the circle at D, E and F respectively. Then,

(a) AD = AB + BC + CA (b) 2AD = AB + BC + CA

(c) 3AD = AB + BC + CA (d) 4AD = AB + BC + CA

3. Let ABC be an acute triangle with circumcenter O, and let K be a point such that KA is tangent to
circle ABC and ∠KCB = 90◦. Point D lies on BC such that KD||AB. Show that line DO passes
through A.

4. A circle has center on the side AB of the cyclic quadrilateral ABCD. The other three sides are tangent
to the circle. Prove that AD + BC = AB.

5. Let ABCD be a non-isosceles trapezium in which AB || CD and AB > CD. Further, ABCD possesses
incentre I, which touches CD at E. Let, M be the mid-point of AB and MI meet CD at F. Show that
DE = FC, if and only if, AB = 2CD.

6. Two circles C1 and C2 of radii 10 cm and 8 cm respectively are tangent to each other internally at a
point A. AD is the diameter of C1 and P and M are points on C1 and C2, as shown in the figure
below. If PM = 20 cm and PAD = x0, find the value of x
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7. Three circles of equal radii(1cm) touch each other. Find the area of triangle formed by direct
tangents of these circle.

8. The tangents at A and B on a given circle O1® intersect at C. Show that the centre of the incircle of
triangle lies on the given circle.

9. As shown in the figure below, circles C1 and C2 of radius 360 are tangent to each other, and both
tangent to straight line l. If circle C3 is tangent of C1, C2 and l and Circe C4 is tangent to C1, C3 and l
find the radius of C4
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10. Let ABCD be a circumscribed (or tangential) quadrilateral. Prove that the circles in the two triangles
ABC and ADC are tangent to each other.

LEVEL - II
11. Let PQ be a chord of a circle and M be the mid-point of PQ. Through M two chords AB and CD of
the circle are drawn. Chords AD and BC intersect PQ at points X and Y respectively. Prove that M is
the mid-point of the segment XY.

12. Let the diagonals of the square ABCD intersect at S and let P be the midpoint of AB. Let
M be the intersection of AC and PD and N the intersection of BD and PC. A circle is
inscribed in the quadrilateral PMSN. Prove that the radius of the circle is MP- MS.

13. Let CD be a chord of a circle 1 and AB a diameter of 1 perpendicular to CD at N with AN > NB.
A circle  2 centred at C with radius CN intersects 1 at points P and Q, and the segments PQ and
CD intersect at M. Given that the radii of 1 and  2 are 61 and 60 respectively, find the length of
AM.

14. The point O is situated inside the parallelogram ABCD such that ∠AOB + ∠COD = 180◦. Prove that
∠OBC = ∠ODC.

15. Coins of the same size are arranging on a very large table (the infinite plane) such that each coin
touches six other coins. Find the percentage of the plane that is covered by the coins

20 50
(a) % (b) % (c) 16 3 % (d) 17 3 %
3 3
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Answers Key
1. (b) 2. (b) 3. (-) 4. (-) 5. (-) 6. (60)
7. (-) 8. (-) 9. (40) 10. (-) 11. (-) 12. (-)
13. (78) 14. (-) 15. (-)

Video Solutions Links

1 https://youtu.be/atGBpQi2c78
2 https://youtu.be/qPAtck0cta4
3 https://youtu.be/HDdoZPf1if8
4 https://youtu.be/QM2pAyI2ycI
5 https://youtu.be/m-PnDJULVYE
6 https://youtu.be/5ZafsBYF9aY
7 https://youtu.be/-nyjTvuJyk0
8 https://youtu.be/peQJonwq_Tk
9 https://youtu.be/HIHkIibBkQk
10 https://youtu.be/jcNdO1u0is0
11 https://youtu.be/MiRjKZs5n1Y
12 https://youtu.be/Rj0X0kJ9LXw
13 https://youtu.be/cU-bVdQd8jY
14 https://youtu.be/zJELi3hFj2o
15 https://youtu.be/ER4TCzqE-W8

Solutions
LEVEL - I
1. In the given figure, three circles with centres A, B, C respectively touch each other externally. If AB
= 5 cm, BC = 7 cm and CA = 6 cm, then the radius of the circle with centre A is
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(a) 1.5 cm (b) 2 cm (c) 2.5 cm (d) 3 cm

Ans. (b)
Sol. Let rA ,rB ,rC are radii of circles A, B and C
AC = rA + rc = 6 …(1)
BC = rB + rC = 7 …(2)
AB = rA + rB = 5 ..(3)
1 + 2 + 3  rA + rB + rC = 9 …. (4)
4 – 2  rA = 2

2. In Fig., if AD, AE and BC are tangents to the circle at D, E and F respectively. Then,

(a) AD = AB + BC + CA (b) 2AD = AB + BC + CA

(c) 3AD = AB + BC + CA (d) 4AD = AB + BC + CA
Ans. (b)
Sol. AD = AE
AD + AE= AC + CD + AB + BE
=AC + CF + AB + BF
2AD = AC + AB + BC

3. Let ABC be an acute triangle with circumcenter O, and let K be a point such that KA is tangent to
circle ABC and ∠KCB = 90◦. Point D lies on BC such that KD||AB. Show that line DO passes
through A.
Sol. Let us assume AD’ passing through O.

∠D’AK + ∠D’CK = 180

D’AKC is cyclic, therefore ∠KD’C = ∠ACK = x
∠ABD’ = ∠CAK = x (Alternate segment theorem)
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∠KD’C = ∠ABD’
⇒ KD’|| AB
But KD||AB ⇒ D’ and D are same.
(M2)

∠KAO =90
∠KDC = ∠CAK
⇒ CDAK is cyclic.
∠DAK + ∠DCK = 180
⇒ ∠DAK = 90
⇒ ∠DAK = ∠KAO =90
A, D, O are collinear.

4. A circle has center on the side AB of the cyclic quadrilateral ABCD. The other three sides are tangent
to the circle. Prove that AD + BC = AB.
Sol.

Let O be the centre of the circle in the problem, and T be the point on AB such that AT = AD. Then
DTA = (180 - DAB)/2 = DCB/2 = DCO
So DCTO is a cyclic quadrilateral
∠ODC = ∠BTC = 90-B/2
∠BCT = 90-B/2
∠BTC = ∠BCT ⇒ BT = BC
AB = AD + BC
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5. Let ABCD be a non-isosceles trapezium in which AB || CD and AB > CD. Further, ABCD possesses
incentre I, which touches CD at E. Let, M be the mid-point of AB and MI meet CD at F. Show that
DE = FC, if and only if, AB = 2CD.
Sol. ∆MIN  ∆FIE

⇒ MN = EF
AB
– BN = CD – ED – FC
2
= CD – 2ED …..(1)
AQ = AN – DE
PB = BN – EC
AD = AG + GD
AD = AN + DE
BC = BH + HC = BN + CE
gr = AD2 – AQ2 = BC2 – BP2
2

4AN.DE = 4BN.CE
AN.DE = BN.CE …..(2)
AN CE
−1 = −1
BN DE
2MN EF
=
BN DE
 BN = 2DE
Put in (1)
AB = 2CD
N = 2DE
CASE II. If AB = 2CD
from (1) BN = 2DE
from (2) AN.DE = BN.CE
AN = 2CE
2MN = AN – BN = 2(CE – DE)
2EF = 2(EF + F(– DE)
FC = DE
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6. Two circles C1 and C2 of radii 10 cm and 8 cm respectively are tangent to each other internally at a
point A. AD is the diameter of C1 and P and M are points on C1 and C2, as shown in the figure
below. If PM = 20 cm and PAD = x0, find the value of x

Ans. 60

Sol.

Let O be the centre of C2 and let PA intersect C2 at B. The homothety centred at A mapping C2 to C1
AB 8
has similitude ratio 8/10. It maps B to P. Thus = . (This can also be seen by connecting P to
AP 10
the centre O’ of C1 so that the triangles ABO and APO’ are similar.) The power of P with respect to
AB 8
C2 is PM2 = 20. Thus PB  PA = 20, or equivalently (PA – AB)PA = 20. Together with = , we
AP 10
obtain AB = 8 and AP = 10. Consequently, the triangle ABO is equilateral, and hence PAD =
BAO = 60°.

7. Three circles of equal radii(1cm) touch each other. Find the area of triangle formed by direct
tangents of these circle.
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Sol.

DA = 2DE = 2
 AE = 3
 Similarly, FB = 3 and EF = CD = 1 + 1 = 2
Hence, side AB = 3 + 3 + 2 = 2(1 + 3)
 Area of equilateral triangle
3
= [2(1 + 3)]2
4

8. The tangents at A and B on a given circle O1® intersect at C. Show that the centre of the incircle of
triangle lies on the given circle.

Sol.
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AO1O2 = 2CAO2 = 2x (Alternate segment theorem)

O1O2 = O1A = R
 O1AO2 = 2AO2O1 = 90 – x
 BAO2 = x
 AO2 is angle bisector of BAC and CO2 is the angle bisector of ABC.
 O2 is the incenter of incircle.

9. As shown in the figure below, circles C1 and C2 of radius 360 are tangent to each other, and both
tangent to straight line l. If circle C3 is tangent of C1, C2 and l and Circe C4 is tangent to C1, C3 and l
find the radius of C4

Ans. 40
Sol. Let R be the radius of C3. Then
(360 – R)2 + 3602 = (360 + R)2  R = 90.

Let r be the radius of C4. Then

= ( 360 + r ) − ( 360 − r ) + ( 90 + r ) − (90 − r ) = 90  r = 40
2 2 2 2
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10. Let ABCD be a circumscribed (or tangential) quadrilateral. Prove that the circles in the two triangles
ABC and ADC are tangent to each other.
Sol. Let the incircle of ∆ABC be C1 and that of ∆ADC be C2

Since C1 and C2 lie on either side of AC, the diagonal, if they touch each other, then, they must touch
at a point only on AC.
If possible let C1 touch AC at P and C2 touch AC at a point Q. (We assume to the contrary).
Then, PQ = AQ – AP (1)
Now AQ = AC – CQ = AC – CR = AC – CD + DR (Equal tangent property)
= AC – CD + DS = AC – CD + DA – SA (Equal tangent property)
= AC – CD + DA – AQ (Equal tangent property)
∴ 2AQ = AC – CD + AD (2)
Similarly, 2AP = AC – BC + AB (3)
∴ 2PQ = (AC – CD + AD) – (AC – BC + AB) (From Eqs. (1), (2) and (3))
i.e., 2PQ = (AD + BC) – (AB + CD) = 0 (by Pitot’s theorem)
Therefore the points P and Q must coincide with each other, i.e., the two circles touch AC at the
same point.

LEVEL - II
11. Let PQ be a chord of a circle and M be the mid-point of PQ. Through M two chords AB and CD of
the circle are drawn. Chords AD and BC intersect PQ at points X and Y respectively. Prove that M is
the mid-point of the segment XY.
Sol. Construction: From X we draw perpendicular lines to AB and CD, with feet X1 and X2
respectively. From Y draw perpendicular lines to AB and CD, with feet Y1 and Y2 respectively.
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let MX = x, MY = y and PM = QM = a.
Using similar triangles we get
x XX1 XX 2 XX1 AX XX 2 DX
= = , = and =
y YY1 YY2 YY2 CY YY1 BY
x 2 XX1 XX 2
 = 
y 2 YY1 YY2
XX1 XX 2
= 
YY2 YY1
AX DX
= 
CY BY
PX  OX
= (By intersecting chords theorem)
PT  QY

=
( a + x )( a − x ) a 2 − x 2
=
( a + y )( a − y ) a 2 − y 2
x2
 =1
y2
x=y
Note: This problem is known as Butterfly theorem.

12. Let the diagonals of the square ABCD intersect at S and let P be the midpoint of AB. Let
M be the intersection of AC and PD and N the intersection of BD and PC. A circle is
inscribed in the quadrilateral PMSN. Prove that the radius of the circle is MP- MS.
Sol. Let O be the centre and r the radius of the circle. Let X, Y be n its points of contact with the sides
PM, MS, respectively
Since OY ⊥ MS and YSO - ASP = 45°, SY = YO = r. Abo OPX = PDA (since OP || DA) and
OXP = PAD = 90°. Therefore OXP PAD . Hence OX/XP = PA/AD = 1/2. Hence Px = 2r.
Therefore PM – MS = 2r + MX - MY – r = r.

13. Let CD be a chord of a circle 1 and AB a diameter of 1 perpendicular to CD at N with AN > NB.
A circle  2 centred at C with radius CN intersects 1 at points P and Q, and the segments PQ and
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CD intersect at M. Given that the radii of 1 and  2 are 61 and 60 respectively, find the length of
AM.
Ans. 78
Sol. Extend DC meeting 2 at H. Note that DN = NC = CH = 60. Since M is of equal power with respect
1 and 2. Thus MN  MH = MC  MD. That is MN(MC + 60) = MC(MN + 60) giving MN = MC.
Thus MN = 30.

The power of N with respect to 1 is DN • NC = 602, and is also equal to NA • NB = NA(AB – NA)
= NA •(122 – NA). Thus NA • (122 – NA) = 602. Solving this quadratic equation, we get NA = 72 or
50. Since NA > NB, we haw NA = 72. Consequently AM = NA2 + MN 2 = 722 + 302 = 78 .

14. The point O is situated inside the parallelogram ABCD such that ∠AOB + ∠COD = 180◦. Prove that
∠OBC = ∠ODC.

Sol.

Let O' be a point such that DAO'O is a parallelogram. Since OO' = DA = BC and all three lines are
parallel, it follows that CBO'O is a parallelogram as well. Moreover, we have AO'B = DOC,
since AO ' || DO and BO ' || CO. Consequently. AO'B + AOB = 180° and AO'BO is cyclic (note
that O' must lie outside the parallelogram since O is given to lie inside it). Actually, one can even
check that O'AB  OBC.
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Consequently. CBO = O'OB = O'AB = ODC as needed.

15. Coins of the same size are arranging on a very large table (the infinite plane) such that each coin
touches six other coins. Find the percentage of the plane that is covered by the coins

20 50
(a) % (b) % (c) 16 3 % (d) 17 3 %
3 3
Ans. (b)
Sol.

Consider the equilateral triangle formed by joining the centres of 3 adjacent coins.
It is easy to see that the required percentage is given by the percentage of the triangle covered by
these three coins.
1 2
r
50
We have 2 2 100% = %.
3r 3