SELINA Solutions for Class 10 Maths Chapter 18 - Tangents and Intersecting Chords
Study easily with Selina Solutions for ICSE Class 10 Mathematics Chapter 18 Tangents and Intersecting Chords. In this chapter, get a deeper understanding of the secant and tangent properties of a circle. TopperLearning’s subject experts will show you how to apply the properties for correctly answering textbook questions with our chapter solutions.
You can further learn about tangents and secants by browsing through our revision notes and ICSE Class 10 Maths videos. To ease your exam anxiety, you can take mock exams using our previous years’ question papers and sample question papers.
Chapter 18 - Tangents and Intersecting Chords Exercise Ex. 18(A)
Question 1
The radius of a circle is 8 cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10 cm from its centre?
Solution 1
OP = 10 cm; radius OT = 8 cm
Length of tangent = 6 cm.
Question 2
In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.
Solution 2
AB = 15 cm, AC = 7.5 cm
Let 'r' be the radius of the circle.
OC = OB = r
AO = AC + OC = 7.5 + r
In ∆AOB,
AO2 = AB2 + OB2
Therefore, r = 11.25 cm
Question 3
Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.
Solution 3
From Q, QA and QP are two tangents to the circle with centre O
Therefore, QA = QP .....(i)
Similarly, from Q, QB and QP are two tangents to the circle with centre O'
Therefore, QB = QP ......(ii)
From (i) and (ii)
QA = QB
Therefore, tangents QA and QB are equal.
Question 4
Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Solution 4
From Q, QA and QP are two tangents to the circle with centre O
Therefore, QA = QP .......(i)
Similarly, from Q, QB and QP are two tangents to the circle with centre O'
Therefore, QB = QP .......(ii)
From (i) and (ii)
QA = QB
Therefore, tangents QA and QB are equal.
Question 5
Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.
Solution 5
OS = 5 cm
OT = 3 cm
In Rt. Triangle OST
By Pythagoras Theorem,
Since OT is perpendicular to SP and OT bisects chord SP
So, SP = 8 cm
Question 6
Three circles touch each other externally. A triangle is formed when the centers of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.
Solution 6
AB = 6 cm, AC = 8 cm and BC = 9 cm
Let radii of the circles having centers A, B and C be r1, r2 and r3 respectively.
r1 + r3 = 8
r3 + r2 = 9
r2 + r1 = 6
Adding
r1 + r3 + r3 + r2 + r2 + r1 = 8+9+6
2(r1 + r2 + r3) = 23
r1 + r2 + r3 = 11.5 cm
r1 + 9 = 11.5 (Since r2 + r3 = 9)
r1 = 2.5 cm
r2 + 6 = 11.5 (Since r1 + r3 = 6)
r2 = 5.5 cm
r3 + 8 = 11.5 (Since r2 + r1 = 8)
r3 = 3.5 cm
Hence, r1 = 2.5 cm, r2 = 5.5 cm and r3 = 3.5 cm
Question 7
If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.
Solution 7
Let the circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively.
Since AP and AS are tangents to the circle from external point A
AP = AS .......(i)
Similarly, we can prove that:
BP = BQ .......(ii)
CR = CQ .......(iii)
DR = DS ........(iv)
Adding,
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence, AB + CD = AD + BC
Question 8
If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.
Solution 8
From A, AP and AS are tangents to the circle.
Therefore, AP = AS.......(i)
Similarly, we can prove that:
BP = BQ .........(ii)
CR = CQ .........(iii)
DR = DS .........(iv)
Adding,
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence, AB + CD = AD + BC
But AB = CD and BC = AD.......(v) Opposite sides of a ||gm
Therefore, AB + AB = BC + BC
2AB = 2 BC
AB = BC ........(vi)
From (v) and (vi)
AB = BC = CD = DA
Hence, ABCD is a rhombus.
Question 9
From the given figure prove that:
AP + BQ + CR = BP + CQ + AR.
Also, show that AP + BQ + CR = 
x perimeter of triangle ABC.
Solution 9
Since from B, BQ and BP are the tangents to the circle
Therefore, BQ = BP ………..(i)
Similarly, we can prove that
AP = AR …………..(ii)
and CR = CQ ………(iii)
Adding,
AP + BQ + CR = BP + CQ + AR ………(iv)
Adding AP + BQ + CR to both sides
2(AP + BQ + CR) = AP + PQ + CQ + QB + AR + CR
2(AP + BQ + CR) = AB + BC + CA
Therefore, AP + BQ + CR = x (AB + BC + CA)
AP + BQ + CR = x perimeter of triangle ABC
Question 10
In the figure, if AB = AC then prove that BQ = CQ.
Solution 10
Since, from A, AP and AR are the tangents to the circle
Therefore, AP = AR
Similarly, we can prove that
BP = BQ and CR = CQ
Adding,
AP + BP + CQ = AR + BQ + CR
(AP + BP) + CQ = (AR + CR) + BQ
AB + CQ = AC + BQ
But AB = AC
Therefore, CQ = BQ or BQ = CQ
Question 11
Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if -
i) they touch each other externally.
ii) they touch each other internally.
Solution 11
Radius of bigger circle = 6.3 cm
and of smaller circle = 3.6 cm
i)
Two circles are touching each other at P externally. O and O’ are the centers of the circles. Join OP and O’P
OP = 6.3 cm, O’P = 3.6 cm
Adding,
OP + O’P = 6.3 + 3.6 = 9.9 cm
ii)
Two circles are touching each other at P internally. O and O’ are the centers of the circles. Join OP and O’P
OP = 6.3 cm, O’P = 3.6 cm
OO’ = OP - O’P = 6.3 - 3.6 = 2.7 cm
Question 12
From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that:
i) 
ii) OP is the perpendicular bisector of chord AB.
Solution 12
i) In 
AP = BP (Tangents from P to the circle)
OP = OP (Common)
OA = OB (Radii of the same circle)
ii) In 
OA = OB (Radii of the same circle)
(Proved )
OM = OM (Common)
Hence, OM or OP is the perpendicular bisector of chord AB.
Question 13
In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that:
i) tangent at point P bisects AB.
ii) Angle APB = 90°
Solution 13
Draw TPT' as common tangent to the circles.
i) TA and TP are the tangents to the circle with centre O.
Therefore, TA = TP ………(i)
Similarly, TP = TB ………..(ii)
From (i) and (ii)
TA = TB
Therefore, TPT' is the bisector of AB.
ii) Now in 
Similarly in 
Adding,
Question 14
Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that:
Solution 14
In quadrilateral OPAQ,
In triangle OPQ,
OP = OQ (Radii of the same circle)
From (i) and (ii)
Question 15
Two parallel tangents of a circle meet a third tangent at point P and Q. Prove that PQ subtends a right angle at the centre.
Solution 15
Join OP, OQ, OA, OB and OC.
In 
OA = OC (Radii of the same circle)
OP = OP (Common)
PA = PC (Tangents from P)
Similarly, we can prove that
(Sum of interior angles of a transversal)
Now in 
Question 16
ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.
Calculate the value of x, the radius of the inscribed circle.
Solution 16
In 
LBNO is a square.
LB = BN = OL = OM = ON = x
Since ABC is a right triangle
Question 17
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate:
i) 
ii) 
given that 
Solution 17
The incircle touches the sides of the triangle ABC and
i) In quadrilateral AROQ,
ii) Now arc RQ subtends at the centre and at the remaining part of the circle.
Question 18
In the following figure, PQ and PR are tangents to the circle, with centre O. If 
, calculate:
i) 
ii) 
iii)
Solution 18
Join QR.
i) In quadrilateral ORPQ,
ii) In 
OQ = QR (Radii of the same circle)
iii) Now arc RQ subtends at the centre and at the remaining part of the circle.
Question 19
In the given figure, AB is a diameter of the circle, with centre O, and AT is a tangent. Calculate the numerical value of x.
Solution 19
In 
OB = OC (Radii of the same circle)
Now in 
Question 20
In quadrilateral ABCD, angle D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm. Find the radius of the circle.
Solution 20
BQ and BR are the tangents from B to the circle.
Therefore, BR =BQ = 27 cm.
Also RC = (38 -; 27) = 11cm
Since CR and CS are the tangents from C to the circle
Therefore, CS = CR = 11 cm
So, DS = (25 - 11) = 14 cm
Now DS and DP are the tangents to the circle
Therefore, DS = DP
Now, 
(given)
and 
therefore, radius = DS = 14 cm
Question 21
In the given figure, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P.
Given 
and 
Prove that -;
i) 
ii) write an expression connecting x and y
Solution 21
(angles in alternate segment)
But OS = OR (Radii of the same circle)
But in 
From (i) and (ii)
Question 22
PT is a tangent to the circle at T. If 
; calculate:
i) 
ii) 
iii) 
Solution 22
Join AT and BT.
i) TC is the diameter of the circle
(Angle in a semi-circle)
ii) 
(Angles in the same segment of the
circle)
(Angles in the same segment of the circle)
iii) 
(Angles in the same segment)
Now in 
Question 23
In the given figure, O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.
Solution 23
Join OC.
Therefore, PA and PA are the tangents
In quadrilateral APCO,
Now, arc BC subtends 
at the centre and 
at the remaining part of the circle
Question 24
In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that : BD is diameter of the circle.
Solution 24
∠CAB = ∠BAQ = 30°……(AB is angle bisector of ∠CAQ)
∠CAQ = 2∠BAQ = 60°……(AB is angle bisector of ∠CAQ)
∠CAQ + ∠PAC = 180°……(angles in linear pair)
∴∠PAC = 120°
∠PAC = 2∠CAD……(AD is angle bisector of ∠PAC)
∠CAD = 60°
Now,
∠CAD + ∠CAB = 60 + 30 = 90°
∠DAB = 90°
Thus, BD subtends 90° on the circle
So, BD is the diameter of circle
Chapter 18 - Tangents and Intersecting Chords Exercise Ex. 18(B)
Question 1
i) In the given figure, 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP.
ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4cm. Find CD.
iii) In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.
Solution 1
i) Since two chords AB and CD intersect each other at P.
ii) Since two chords AB and CD intersect each other at P.
iii) Since PAB is the secant and PT is the tangent
Question 2
In the given figure, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find
(i) AB.
(ii) the length of tangent PT.
Solution 2
Question 3
In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If 
; calculate:
i) 
ii) 
iii) 
Solution 3
i) PAQ is a tangent and AB is the chord.
(angles in the alternate segment)
ii) OA = OD (radii of the same circle)
iii) BD is the diameter.
(angle in a semi-circle)
Now in 
Question 4
If PQ is a tangent to the circle at R; calculate:
i) 
ii) 
Given: O is the centre
of the circle and 
Solution 4
PQ is a tangent and OR is the radius.
But in 
OT = OR (Radii of the same circle)
In 
Question 5
AB is diameter and AC is a chord of a circle with centre O such that angle BAC=30º. The tangent to the circle at C intersects AB produced in D. Show that BC = BD.
Solution 5
Join OC.
(angles in alternate segment)
Arc BC subtends 
at the centre of the circle and 
at the remaining part of the circle.
Now in 
Now in 
Question 6
Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that triangle PQR is isosceles.
Solution 6
DE is the tangent to the circle at P.
DE||QR (Given)
Since the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment
(DE is tangent and PQ is
chord)
from (i) and (ii)
Hence, triangle PQR is an isosceles triangle.
Question 7
Two circles with centers O and O' are drawn to intersect each other at points A and B.
Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O' at A. Prove that OA bisects angle BAC.
Solution 7
Join OA, OB, O'A, O'B and O'O.
CD is the tangent and AO is the chord.
(angles in alternate
segment)
In 
OA = OB (Radii of the same circle)
From (i) and (ii)
Therefore, OA is bisector of 
BAC
Question 8
Two circles touch each other
internally at a point P. A chord AB of the bigger circle intersects the other
circle in C and D. Prove that: 
Solution 8
Draw a tangent TS at P to the circles given.
Since TPS is the tangent, PD is the chord.
Subtracting (i) from (ii)
But in 
Question 9
In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.
Solution 9
TAS is a tangent and AB is a chord
But these are alternate angles
Therefore, TS||BD.
Question 10
In the figure, ABCD is a cyclic
quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is
produced to point G. If angle BCG=108
and O is the centre
of the circle, find:
i) angle BCT
ii) angle DOC
Solution 10
Join OC, OD and AC.
i)
ii)
PCT is a tangent and CA is a chord.
But arc DC subtends 
at the centre and 
at the
remaining part of the circle.
Question 11
Two circles intersect each other at point A and B. A straight line PAQ cuts the circle at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T are concyclic.
Solution 11
Join AB, PB and BQ
TP is the tangent and PA is a chord
(angles in alternate segment)
Similarly,
Adding (i) and (ii)
But they are the opposite angles of the quadrilateral
Therefore, PBQT are cyclic.
Hence, P, B, Q and T are concyclic.
Question 12
In the figure, PA is a tangent to the circle. PBC is a secant and AD bisects angle BAC.
Show that the triangle PAD is an isosceles triangle. Also show that:
Solution 12
i) PA is the tangent and AB is a chord
( angles in the alternate
segment)
AD is the bisector of 
In 
Therefore, 
is an isosceles
triangle.
ii) In 
Question 13
Two circles intersect each other at point A and B. Their common tangent touches the circles at points P and Q as shown in the figure. Show that the angles PAQ and PBQ are supplementary.
Solution 13
Join AB.
PQ is the tangent and AB is a chord
(angles in alternate segment)
Similarly,
Adding (i) and (ii)
From (iii) and (iv)
Hence, 
and 
are supplementary.
Question 14
In the figure, chords AE and BC intersect each other at point D.
i) if 
, AB = 5 cm, BD = 4 cm and CD = 9 cm; find DE
ii) If AD = BD, Show that AE = BC.
Solution 14
Join AB.
i) In Rt. 
Chords AE and CB intersect each other at D inside the circle
AD x DE = BD x DC
3 x DE = 4 x 9
DE = 12 cm
ii) If AD = BD .......(i)
We know that:
AD x DE = BD x DC
But AD = BD
Therefore, DE = DC .......(ii)
Adding (i) and (ii)
AD + DE = BD + DC
Therefore, AE = BC
Question 15
Circles with centers P and Q intersect at points A and B as shown in the figure. CBD is a line segment and EBM is tangent to the circle, with centre Q, at point B. If the circles are congruent; show that CE = BD.
Solution 15
Join AB and AD
EBM is a tangent and BD is a chord.
(angles in alternate segments)
(Vertically opposite angles)
Since in the same circle or congruent circles, if angles are equal, then chords opposite to them are also equal.
Therefore, CE = BD
Question 16
In the adjoining figure, O is the
centre of the circle and AB is a tangent to it at point B. 
Find 
Solution 16
AB is a straight line.
AB i.e. DB is tangent to the circle at point B and BC is the diameter.
Now, OE = OC (radii of the same circle)
(vertically opposite angles)
In 
Chapter 18 - Tangents and Intersecting Chords Exercise Ex. 18(C)
Question 1
Prove that of any two chord of a circle, the greater chord is nearer to the centre.
Solution 1
Given: A circle with centre O and
radius r. 
. Also AB > CD
To prove: OM < ON
Proof: Join OA and OC.
In Rt.
Again in Rt. 
From (i) and (ii)
Hence, AB is nearer to the centre than CD.
Question 2
OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.
i)If the radius of the circle is 10 cm, find the area of the rhombus.
ii) If the area of the rhombus is 
, find the radius of
the circle.
Solution 2
i) Radius = 10 cm
In rhombus OABC,
OC = 10 cm
In Rt.
Area of rhombus = 
ii) Area of rhombus =
But area of rhombus OABC = 2 x
area of 
Where r is the side of the equilateral triangle OAB.
Therefore, radius of the circle = 8 cm
Question 3
Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.
Solution 3
If two circles touch internally, then distance between their centres is equal to the difference of their radii. So, AB = (5 - 3) cm = 2 cm.
Also, the common chord PQ is the perpendicular bisector of AB. Therefore, AC = CB = 
AB = 1 cm
In right 
ACP, we have AP2 = AC2 + CP2
52 = 12 + CP2
CP2 = 25 -; 1 = 24
CP = 
Now, PQ = 2 CP
= 2 x 
cm
Question 4
Two chords AB and AC of a circle are equal. Prove that the centre of the circle, lies on the bisector of the angle BAC.
Solution 4
Given: AB and AC are two equal chords of C (O, r).
To prove: Centre, O lies on the bisector of 
BAC.
Construction: Join BC. Let the bisector of BAC intersects BC in P.
Proof:
In 
APB and APC,
AB = AC (Given)
BAP = CAP (Given)
AP = AP (Common)
(SAS congruence criterion)
BP = CP and APB = APC (CPCT)
APB + APC = 180
(Linear pair)
=> 2APB = 180 (APB = APC)
APB = 90
Now, BP = CP and APB = 90
AP is the perpendicular bisector of chord BC.
AP passes through the centre, O of the circle.
Question 5
The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the centre of the circle?
Solution 5
AB is the diameter and AC is the chord.
Draw 
Since and hence it bisects AC, O is the centre of the circle.
Therefore, OA = 10 cm and AL = 6 cm
Now, in Rt.
Therefore, chord is at a distance of 8 cm from the centre of the circle.
Question 6
ABCD is a cyclic quadrilateral in
which BC is parallel to AD, angle ADC = 
and angle BAC = 
. Find angle DAC and
angle DCA.
Solution 6
ABCD is a cyclic quadrilateral in which AD||BC
(Sum of opposite angles of a quadrilateral)
Now in 
Now in 
Question 7
In the given figure, C and D are points on the semicircle described on AB as diameter.
Given angle BAD = 70
and angle DBC = 30, calculate angle BDC
Solution 7
Since ABCD is a cyclic quadrilateral, therefore, 
BCD + BAD = 180
(since opposite angles of a cyclic quadrilateral are supplementary)
BCD + 70 = 180
BCD = 180 - 70 = 110
In 
BCD, we have,
CBD + BCD + BDC = 180
30 + 110 + BDC = 180
BDC = 180 - 140
BDC = 40
Question 8
In cyclic quadrilateral ABCD, 
A = 3C and D = 5B. Find the measure of
each angle of the quadrilateral.
Solution 8
ABCD is a cyclic quadrilateral.
Similarly,
Hence, 
Question 9
Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution 9
Join AD.
AB is the diameter.
ADB = 90º (Angle in a semi-circle)
But, 
ADB + ADC = 180º (linear pair)
ADC = 90º
In 
ABD and ACD,
ADB = ADC (each 90º)
AB = AC (Given)
AD = AD (Common)
ABD 
ACD (RHS congruence criterion)
BD = DC (C.P.C.T)
Hence, the circle bisects base BC at D.
Question 10
Bisectors of vertex A, B and C of
a triangle ABC intersect its circumcircle at points D, E and F respectively.
Prove that angle EDF =
Solution 10
Join ED, EF and DF. Also join BF, FA, AE and EC.
In cyclic quadrilateral AFBE,
(Sum of opposite angles)
Similarly in cyclic quadrilateral CEAF,
Adding (ii) and (iii)
Question 11
In the figure, AB is the chord of
a circle with centre O and DOC is a line segment such that BC = DO. If 
, find angle AOD.
Solution 11
Join OB.
In 
In 
Since DOC is a straight line
Question 12
Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
Solution 12
Join OL, OM and ON.
Let D and d be the diameter of the circumcircle and incircle.
and let R and r be the radius of the circumcircle and incircle.
In circumcircle of 
Therefore, AC is the diameter of the circumcircle i.e. AC = D
Let radius of the incircle = r
Now, from B, BL, BM are the tangents to the incircle.
(Tangents from the point outside the circle)
Now,
AB+BC+CA = AM+BM+BL+CL+CA
= AN+r+r+CN+CA
= AN+CN+2r+CA
= AC+AC+2r
= 2AC+2r
= 2D+d
Question 13
P is the midpoint of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.
Solution 13
Join AP and BP.
Since TPS is a tangent and PA is the chord of the circle.
(angles in alternate segments)
But
But these are alternate angles
Question 14
In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent.
Prove that the line NM produced bisects AB at P.
Solution 14
From P, AP is the tangent and PMN is the secant for first circle.
Again from P, PB is the tangent and PMN is the secant for second circle.
From (i) and (ii)
Therefore, P is the midpoint of AB.
Question 15
In the given figure, ABCD is a
cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its
diameter. If 
and 
, find:
i) 
ii) 
iii) 
Solution 15
i) PQ is tangent and CD is a chord
(angles in the alternate segment)
ii)
iii) In 
Question 16
The given figure shows a circle
with centre O and BCD is a tangent to it at C. Show that: 
Solution 16
Join OC.
BCD is the tangent and OC is the radius.
Substituting in (i)
Question 17
ABC is a right triangle with angle B = 90º. A circle with BC as diameter meets by hypotenuse AC at point D.
Prove that -
i) AC x AD = AB2
ii) BD2 = AD x DC.
Solution 17
i) In 
and BC is the diameter of the circle.
Therefore, AB is the tangent to the circle at B.
Now, AB is tangent and ADC is the secant
ii) In 
From (i) and (ii)
Now in 
Question 18
In the given figure AC = AE.
Show that:
i) CP = EP
ii) BP = DP
Solution 18
In 
(angles in the same segment)
AC = AE (Given)
(Common)
(ASA Postulate)
AB = AD
but AC = AE
In 
(angles in the same segment)
BC = DE
(angles in the same segment)
(ASA Postulate)
BP = DP and CP = PE (cpct)
Question 19
ABCDE is a cyclic pentagon with
centre of its circumcircle at point O such that AB = BC = CD and angle
ABC=120
Calculate:
i) 
ii)
Solution 19
i) Join OC and OB.
AB = BC = CD and 
OB and OC are the bisectors of 
and 
respectively.
In 
Arc BC subtends 
at the centre and at the remaining part
of the circle.
ii) In cyclic quadrilateral BCDE,
Question 20
In the given figure, O is the
centre of the circle. Tangents at A and B meet at C. If angle ACO = 30
, find:
(i) angle BCO
(ii) angle AOB
(iii) angle APB
Solution 20
In the given fig, O is the centre
of the circle and CA and CB are the tangents to the circle from C. Also, 
ACO = 30
P is any point on the circle. P and PB are joined.
To find: (i) 
(ii) 
(iii)
Proof:
Question 21
ABC is a triangle with AB = 10 cm, BC = 8 cm and AC = 6cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles.
Solution 21
Given: ABC is a triangle with AB = 10 cm, BC= 8 cm, AC = 6 cm. Three circles are drawn with centre A, B and C touch each other at P, Q and R respectively.
We need to find the radii of the three circles.
Question 22
In a square ABCD, its diagonal AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and bisector of angle ABD meets AC at N and AM at L. Show that -
i) 
ii) 
iii) ALOB is a cyclic quadrilateral.
Solution 22
ABCD is a square whose diagonals AC and BD intersect each other at right angles at O.
i) 
In 
But, 
(vertically opposite angles)
Now in 
Adding (i) and (ii)
ii)
and
iii) In quadrilateral ALOB,
Therefore, ALOB is a cyclic quadrilateral.
Question 23
The given figure shows a
semicircle with centre O and diameter PQ. If PA = AB and 
; find measures of
angles PAB and AQB. Also, show that AO is parallel to BQ.
Solution 23
Join PB.
i) In cyclic quadrilateral PBCQ,
Now in 
In cyclic quadrilateral PQBA,
ii) Now in 
iii) Arc AQ subtends 
at the centre and 
APQ at the remaining
part of the circle.
We have,
From (1), (2) and (3), we have
Now in 
But these are alternate angles.
Hence, AO is parallel to BQ.
Question 24
The given figure shows a circle with centre O such that chord RS is parallel to chord QT, angle PRT = 20
and angle POQ = 100.
Calculate -
i) angle QTR
ii) angle QRP
iii) angle QRS
iv) angle STR
Solution 24
Join PQ, RQ and ST.
i)
Arc RQ subtends 
at the centre and 
QTR at the remaining part of the circle.
ii) Arc QP subtends 
at the centre and QRP at the remaining part of the circle.
iii) RS || QT
iv) Since RSTQ is a cyclic quadrilateral
(sum of opposite angles)
Question 25
In the given figure, PAT is tangent to the circle with centre O, at point A on its circumference and is parallel to chord BC. If CDQ is a line segment, show that:
Solution 25
i) Since PAT||BC
(alternate angles) .........(i)
In cyclic quadrilateral ABCD,
from (i) and (ii)
ii) Arc AB subtends 
at the centre and 
at the remaining part of the circle.
iii)
Question 26
AB is a line segment and M is its midpoint. Three semicircles are drawn with AM, MB and AB as diameters on the same side of the line AB. A circle with radius r unit is drawn so that it touches all the three semicircles. Show that: AB = 6 x r
Solution 26
Let O, P and Q be the centers of the circle and semicircles.
Join OP and OQ.
OR = OS = r
and AP = PM = MQ = QB = 
Now, OP = OR + RP = r + (since PM=RP=radii of
same circle)
Similarly, OQ = OS + SQ = r +
OM = LM -; OL = 
- r
Now in Rt. 
Hence AB = 6 x r
Question 27
TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Prove that AP bisects the angle TAB.
Solution 27
Join PB.
In 
TAP and TBP,
TA = TB (tangents segments from an external points are equal in length)
Also, 
ATP = BTP. (since OT is equally inclined with TA and TB) TP = TP (common)
TAP 
TBP (by SAS criterion of congruency)
TAP = TBP (corresponding parts of congruent triangles are equal)
But TBP = BAP (angles in alternate segments)
Therefore, TAP = BAP.
Hence, AP bisects TAB.
Question 28
Two circles intersect in points P and Q. A secant passing through P intersects the circle in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.
Solution 28
Join PQ.
AT is tangent and AP is a chord.
(angles in alternate segments) ........(i)
Similarly, 
.......(ii)
Adding (i) and (ii)
Now in 
Therefore, AQBT is a cyclic quadrilateral.
Hence, A, Q, B and T lie on a circle.
Question 29
Prove that any four vertices of a regular pentagon are concyclic (lie on the same circle)
Solution 29
ABCDE is a regular pentagon.
In 
AED,
AE = ED (Sides of regular pentagon ABCDE)
EAD = EDA
In AED,
AED + EAD + EDA = 180º
108º + EAD + EAD = 180º
2EAD = 180º - 108º = 72º
EAD = 36º
EDA = 36º
BAD = BAE - EAD = 108º - 36º = 72º
In quadrilateral ABCD,
BAD + BCD = 108º + 72º = 180º
ABCD is a cyclic quadrilateral
Question 30
Chords AB and CD of a circle when extended meet at point X. Given AB = 4 cm, BX = 6 cm and XD = 5 cm. Calculate the length of CD.
Solution 30
We know that XB.XA = XD.XC
Or, XB.(XB + BA) = XD.(XD + CD)
Or, 6(6 + 4) = 5(5 + CD)
Or, 60 = 5(5 + CD)
Or, 5 + CD = 
= 12
Or, CD = 12 - 5 = 7 cm.
Question 31
In the given figure, find TP if AT = 16 cm and AB = 12 cm.
Solution 31
PT is the tangent and TBA is the secant of the circle.
Therefore, TP2 = TA x TB
TP2 = 16 x (16-12) = 16 x 4 = 64 = (8)2
Therefore, TP = 8 cm
Question 32
In the following figure, A circle is inscribed in the quadrilateral ABCD.
If BC = 38 cm, QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle.
Solution 32
From the figure we see that BQ = BR = 27 cm (since length of the tangent segments from an
external point are equal)
As BC = 38 cm
CR = CB - BR = 38 - 27
= 11 cm
Again,
CR = CS = 11cm (length of tangent segments from an external point are equal)
Now, as DC = 25 cm
DS = DC - SC
= 25 -11
= 14 cm
Now, in quadrilateral DSOP,
PDS = 90
(given)
OSD = 90, OPD = 90 (since tangent is perpendicular to the
radius through the point of contact)
DSOP is a parallelogram
OP||SD and PD||OS
Now, as OP = OS (radii of the same circle)
OPDS is a square. DS = OP = 14cm
radius of the circle = 14 cm
Question 33
In the figure, XY is the diameter of the circle, PQ is the tangent to the circle at Y. Given that 
and 
. Calculate 
and 
.
Solution 33
In 
AXB,
XAB + AXB + ABX=180
[Triangle property]
XAB + 50 + 70 = 180
XAB=180 - 120 = 60
XAY=90 [Angle of semi-circle]
BAY=XAY -XAB = 90 - 60 = 30
and BXY = BAY = 30 [Angle of same segment]
ACX = BXY + ABX [External angle = Sum of two interior angles]
= 30 + 70
= 100
also,
XYP=90 [Diameter ⊥ tangent]
APY = ACX -CYP
APY=100 - 90
APY=10
Question 34
In the given figure, QAP is the
tangent at point A and PBD is a straight line. If 
and 
; find:
i) 
ii)
iii) 
iv) 
Solution 34
PAQ is a tangent and AB is a chord of the circle.
i) 
(angles in alternate segment)
ii) In 
iii) 
(angles in the same
segment)
Now in 
iv) PAQ is the tangent and AD is chord
Question 35
In the given figure, AB is the diameter. The tangent at C meets AB produced at Q.
If
Solution 35
i) AB is diameter of circle.
In 
ii) QC is tangent to the circle
Angle between tangent and chord = angle in alternate segment
ABQ is a straight line
Question 36
In the given figure, O is the centre of the circle. The tangets at B and D intersect each other at point P.
If AB is parallel to CD and 
, find:
i) 
ii) 
Solution 36
i)
ii) Since, BPDO is cyclic quadrilateral, opposite angles are supplementary.
Question 37
In the figure given below PQ =QR, 
RQP = 68
, PC and CQ are
tangents to the circle with centre O. Calculate the values of:
i) QOP
ii) QCP
Solution 37
i) PQ = RQ
(opposite angles of equal sides
of a triangle)
Now, QOP = 2PRQ (angle at the
centre is double)
ii) PQC = PRQ (angles in
alternate segments are equal)
QPC = PRQ (angles in
alternate segments)
Question 38
In two concentric circles prove that all chords of the outer circle, which touch the inner circle, are of equal length.
Solution 38
Consider two concentric circles with centres at O. Let AB and CD be two chords of the outer circle which touch the inner circle at the points M and N respectively.
To prove the given question, it is sufficient to prove AB = CD.
For this join OM, ON, OB and OD.
Let the radius of outer and inner circles be R and r respectively.
AB touches the inner circle at M.
AB is a tangent to the inner
circle
OM
AB
BM = 
AB
AB = 2BM
Similarly ONCD, and CD = 2DN
Using Pythagoras
theorem in 
OMB and OND
Question 39
In the figure, given below, AC is a transverse common tangent to two circles with centers P and Q and of radii 6 cm and 3 cm respectively.
Given that AB = 8 cm, calculate PQ.
Solution 39
Since AC is tangent to the circle with center P at point A.
In 
Also in Rt.
From (i) and (ii),
Question 40
In the figure given below, O is
the centre of the circum circle of triangle XYZ. Tangents at X and Y
intersect at point T. Given 
XTY
= 80
and XOZ = 140, calculate the value
of ZXY.
Solution 40
In the figure, a circle with centre O, is the circum circle of triangle XYZ.
Tangents at X and Y intersect at
point T, such that XTY = 80
Question 41
In the given figure, AE and BC
intersect each other at point D. If 
CDE=90
, AB = 5 cm, BD = 4 cm
and CD = 9 cm, find AE.
Solution 41
From Rt. 
Now, since the two chords AE and BC intersect at D,
AD x DE = CD x DB
3 x DE = 9 x 4
Hence, AE = AD + DE = (3 + 12) = 15 cm
Question 42
In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.
Solution 42
Question 43
In the figure given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x, y and z.
Solution 43
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change