SECANTS, TANGENTS AND THEIR PROPERTIES -GEOMETRY

Mathematics Secondary Course 409
Secants, Tangents and Their Properties
Notes
MODULE - 3
Geometry
17
SECANTS, TANGENTS AND THEIR
PROPERTIES
Look at the moving cycle.You will observe that at any instant of time, the wheels of the
moving cycle touch the road at a very limited area, more correctly a point.
If you roll a coin on a smooth surface, say a table or floor, you will find that at any instant
of time, only one point of the coin comes in contact with the surface it is rolled upon.
What do you observe from the above situations?
(i) (ii)
(iii)
Fig. 17.1
Fotball

Mathematics Secondary Course410
Notes
MODULE - 3
Geometry
If you consider a wheel or a coin as a circle and the touching surface (road or table) as a
line,theaboveillustrationsshowthatalinetouchesacircle.Inthislesson,weshallstudy
aboutthepossiblecontactsthatalineandacirclecanhaveandtrytostudytheirproperties.
OBJECTIVES
Afterstudyingthislesson,youwillbeableto
• define a secant and a tangent to the circle;
• differentitate between a secant and a tangent;
• prove that the tangents drawn from an external point to a circle are of equal
length;
• verify the un-starred results (given in the curriculum) related to tangents and
secants to circle experimentally.
EXPECTED BACKGROUND KNOWLEDGE
• Measurementofanglesandlinesegments
• Drawingcirclesofgivenradii
• Drawinglinesperpendicularandparalleltogivenlines
• Knowledgeofpreviousresultsaboutlinesandangles,congruenceandcircles
• KnowledgeofPythagorasTheorem
17.1 SECANTS AND TANGENTS—AN INTRODUCTION
Youhavereadaboutlinesandcirclesinyourearlierlessons.Recallthatacircleisthelocus
of a point in a plane which moves in such a way that its distance from a fixed point in the
plane always remains constant. The fixed point is called the centre of the circle and the
constantdistanceiscalledtheradiusofthecircle.Youalsoknowthatalineisacollection
ofpoints,extendingindefinitelytobothsides,whereasalinesegmentisaportionofaline
bounded by two points.
(i) (ii) (iii)
Fig. 17.2

Notes
MODULE - 3
Geometry
Now consider the case when a line and a circle co-exist in the same plane. There can be
threedistinctpossibilitiesasshowninFig.17.2.
You can see that in Fig. 17.2(i), the XY does not intersect the circle, with centre O. In
other words, we say that the line XY and the circle have no common point. In Fig. 17.2
(ii),thelineXYintersectsthecircleintwodistinctpointAandB,andinFig.17.2(iii),the
line XYintersects the circle in only one point and is said to touch the circle at the point P.
Thus, we can say that in case of intersection of a line and a circle, the following three
possibilitiesarethere:
(i) Thelinedoesnotintersectthecircleatall,i.e.,thelineliesintheexteriorofthecircle.
(ii) Thelineintersectsthecircleattwodistinctpoints.Inthatcase,apartofthelineliesin
theinteriorofthecircle,thetwopointsofintersectionlieonthecircleandtheremaining
portionofthelineliesintheexteriorofthecircle.
(iii) Thelinetouchesthecircleinexactlyonepoint.Wethereforedefinethefollowing:
Tangent:
A line which touches a circle at exactly one point is called a tangent line and
the point where it touches the circle is called the point of contact
Thus,inFig.17.2(iii),XYisatangentofthecircleatP,whichiscalledthepointofcontact.
Secant:
A line which interesects the circle in two distinct points is called a secant
line (usually referred to as a secant).
In Fig. 17.2 (ii), XY is a secant line to the circle and A and B are called the points of
intersection of the line XY and the circle with centre O.
17.2 TANGENT AS A LIMITING CASE
Consider the secant XY of the circle with centre
O, intersecting the circle in the pointsAand B.
ImaginethatonepointA,whichliesonthecircle,
of the secant XY is fixed and the secant rotates
about A, intersecting the circle at B′, B′′, B′′′,
B′′′′ as shown in Fig. 17.3 and ultimately attains
the position of the line XAY, when it becomes
tangent to the circle atA.
Thus,wesaythat:
A tangent is the limiting position of a
secant when the two points of intersection
coincide.
Fig. 17.3

Notes
MODULE - 3
Geometry
17.3 TANGENT AND RADIUS THROUGH THE POINT OF
CONTACT
Let XYbe a tangent to the circle, with centre O, at the point P. Join OP.
Take points Q, R, S and T on the tangent XY and join OQ, OR, OS and OT.
As Q, R, S and T are points in the exterior of the circle and P is on the circle.
∴ OP is less than each of OQ, OR, OS and OT.
Fromour,“previousstudyofGeometry,weknow
that of all the segments that can be drawn from a
point(notontheline)totheline,theperpendicular
segmentistheshortest”:
As OP is the shortest distance from O to the line
XY
∴ OP ⊥ XY
Thus, we can state that
A radius, though the point of contact of tangent to a circle, is perpendicular
to the tangent at that point.
The above result can also be verified by measuring angles OPX and OPY and
finding each of them equal to 90o
.
17.4 TANGENTS FROM A POINT OUTSIDE THE CIRCLE
Take any point P in the exterior of the circle with
centre O. Draw lines through P. Some of these are
shown as PT, PA, PB, PC, PD and PT′ in Fig. 17.5
How many of these touch the circle? Only two.
Repeattheactivitywithanotherpointandacircle.
Youwillagainfindthesameresult.
Thus, we can say that
From an external point, two tangents can
be drawn to a circle.
IfthepointPliesonthecircle,cantherestillbetwotangentstothecirclefromthatpoint?
You can see that only one tangent can be drawn to the circle in that case. What about the
case when P lies in the interior of the circle? Note that any line through P in that case will
intersectthecircleintwopointsandhencenotangentcanbedrawnfromaninteriorpoint
to the circle.
(A) Now, measure the lengths of PTand PT′.You will find that
PT = PT′ ....(i)
Fig. 17.4
Fig. 17.5

Notes
MODULE - 3
Geometry
(B) Given: AcirclewithcentreO.PTandPT′ aretwotangentsfromapointPoutsidethe
circle.
To Prove: PT = PT′
Construciton: Join OP, OT and OT′ (see Fig. 17.6)
Proof: In Δ’s OPT and OPT′
∠ OTP = ∠ OT′P (Each being right angle)
OT = OT′
OP = OP (Common)
ΔOPT ≅ ΔOPT′ (RHS criterion)
∴ PT = PT′
The lengths of two tangents from an external point are equal
Also, from Fig. 17.6, ∠ OPT = ∠ OPT′ (As ΔOPT ≅ ΔOPT′)
The tangents drawn from an external point to a circle are equally inclined to
the line joining the point to the centre of the circle.
Let us now take some examples to illustrate:
Example 17.1: InFig.17.7,OP=5cmandradiusofthecircleis3cm.Findthelengthof
the tangent PT from P to the circle, with centre O.
Solution: ∠OTP = 90o
, Let PT = x
InrighttriangleOTP,wehave
OP2
= OT2
+ PT2
or 52
= 32
+ x2
or x2
= 25 – 9 = 16
∴ x = 4
i.e. the length of tangent PT = 4 cm
Example 17.2: In Fig. 17.8, tangents PT and PT′ are drawn from a point P at a distance
of 25 cm from the centre of the circle whose radius is 7 cm. Find the lengths of PT and
PT′.
Solution: Here OP = 25 cm and OT = 7 cm
We also know that
∠OTP = 90o
∴ PT2
= OP2
– OT2
= 625 – 49 = 576 = (24)2
∴ PT = 24 cm
Fig. 17.6
Fig. 17.8
Fig. 17.7

Notes
MODULE - 3
Geometry
We also know that
PT = PT′
∴ PT′ = 24 cm
Example 17.3: In Fig. 17.9,A, B and C are three exterior points of the circle with centre
O. The tangentsAP, BQ and CR are of lengths 3 cm, 4 cm and 3.5 cm respectively. Find
the perimeter of ΔABC.
Solution: We know that the lengths of two tangents from an external point to a circle are
equal
∴ AP =AR
BP = BQ,
CQ = CR
∴ AP = AR = 3 cm
BP = BQ = 4 cm
and CR = CQ = 3.5 cm
AB = AP + PB;
= (3 + 4) cm = 7 cm
BC = BQ + QC;
= (4 + 3.5) cm = 7.5 cm
CA = AR + CR
= (3 + 3.5) cm
∴ = 6.5 cm
∴ Perimeter of ΔABC = (7 + 7.5 + 6.5) cm = 21 cm
Example 17.4: In Fig. 17.10, ∠AOB = 50o
. Find ∠ABO and ∠OBT.
Solution: We know that OA ⊥ XY
⇒ ∠OAB = 90o
∴ ∠ABO = 180o
– (∠OAB + ∠AOB)
= 180o
– (90o
+ 50o
) = 40o
We know that ∠OAB = ∠OBT
⇒ ∠OBT = 40o
∴ ∠ABO = ∠OBT = 40o
Fig. 17.9
Fig. 17.10
B

Notes
MODULE - 3
Geometry
CHECK YOUR PROGRESS 17.1
1. Fillintheblanks:
(i) A tangent is __________ to the radius through the point of contact.
(ii) The lengths of tangents from an external point to a circle are __________
(iii) Atangentisthelimitingpositionofasecantwhenthetwo______coincide.
(iv) From an external point ________ tangents can be drawn to a circle.
(v) From a point in the interior of the circle, ______ tangent(s) can be drawn to the
circle.
2. In Fig. 17.11, ∠POY = 40o
, Find the ∠OYP and ∠OYT.
3. In Fig. 17.12, the incircle of ΔPQR is drawn. If PX = 2.5 cm, RZ = 3.5 cm and
perimeter of ΔPQR = 18 cm, find the lenght of QY.
Fig. 17.11 Fig. 17.12
4. Write an experiment to show that the lengths of tangents from an external point to a
circle are equal.
17.5 INTERSECTING CHORDS INSIDE AND OUTSIDE A
CIRCLE
Youhavereadvariousresultsaboutchordsintheprevious
lessons.Wewillnowverifysomeresultsregardingchords
intersecting inside a circle or outside a circle, when
produced.
Letusperformthefollowingactivity:
Draw a circle with centre O and any radius. Draw two
chordsAB and CD intersecting at Pinside the circle.
Measure the lenghts of the line-segments PD, PC, PA
and PB. Find the products PA × PB and PC × PD.
Fig. 17.13

Notes
MODULE - 3
Geometry
Youwillfindthattheyareequal.
Repeattheaboveactivitywithanothercircleafterdrawingchrodsintersectinginside.You
willagainfindthat
PA × PB = PC × PD
Let us now consider the case of chrods intersecting outside the circle. Let us perform the
followingactivity:
Draw a circle of any radius and centre O. Draw two chords BA and DC intersecting each
other outside the circle at P. Measure the lengths of line segments PA, PB, PC and PD.
Find the products PA× PB and PC × PD.
YouwillseethattheproductPA×PBisequaltothe
product PC × PD, i.e.,
PA × PB = PC × PD
Repeat this activity with two circles with chords
intersectingoutsidethecircle.Youwillagainfindthat
PA × PB = PC × PD.
Thus, we can say that
If two chords AB and CD of a circle intersect at a point P (inside or outside the
circle), then
PA × PB = PC × PD
17.6 INTERSECTING SECANT AND TANGENT OF A
CIRCLE
Toseeifthereissomerelationbeweentheintersecting
secant and tangent outside a circle, we conduct the
followingactivity.
Draw a circle of any radius with centre O. From an
external point P, draw a secant PAB and a tangent
PT to the circle.
Measure the length of the line-segment PA, PB and
PT. Find the products PA × PB and PT × PT or
PT2
. What do you find?
Youwillfindthat
PA × PB = PT2
Repeattheaboveactivitywithtwoothercircles.Youwillagainfindthesameresult.
Fig. 17.14
Fig. 17.15

Notes
MODULE - 3
Geometry
Thus, we can say
If PAB is a secant to a circle intersecting the circle at A and B, and PT is a
tangent to the circle at T, then
PA × PB = PT2
Letusillustratethesewiththehelpofexamples:
Example 17.5: In Fig. 17.16,AB and CD are two
chords of a circle intersecting at a point P inside the
circle. If PA = 3 cm, PB = 2 cm and PC = 1.5 cm,
thenfindthelengthofPD.
Solution: It is given that PA= 3 cm, PB = 2 cm and
PC = 1.5 cm.
Let PD = x
We know that PA × PB = PC × PD
⇒ 3 × 2 = (1.5) × x
⇒ 4
1.5
23
x =
×
=
∴ Length of the line-segment PD = 4 cm.
Example 17.6: In Fig. 17.17, PAB is a secant to the circle from a point P outside the
circle. PAB passes through the centre of the circle and PT is a tangent. If PT = 8 cm and
OP= 10 cm, find the radius of the circle, using PA× PB = PT2
Solution: Let x be the radius of the circle.
Itisgiventhat OP = 10 cm
∴ PA = PO – OA = (10 – x) cm
and PB = OP + OB = (10 + x) cm
PT = 8 cm
We know that PA × PB = PT2
∴ (10 – x) (10 + x) = 82
or 100 – x2
= 64
or x2
= 36 or x = 6
i.e., radius of the circle is 6 cm.
Example 17.7: In Fig. 17.18, BA and DC are two chords of a circle intersecting each
other at a point P outside the circle. If PA= 4 cm, PB = 10 cm, CD = 3 cm, find PC.
Fig. 17.16
Fig. 17.17

Notes
MODULE - 3
Geometry
Solution: We are given that PA= 4 cm, PB = 10 cm, CD = 3 cm
Let PC =x
We know that PA × PB = PC × PD
or 4 × 10 = (x + 3) x
or x2
+ 3x – 40 = 0
(x + 8) (x – 5) = 0
⇒ x = 5
∴ PC = 5 cm
1. In Fig. 17.19, if PA= 3 cm, PB = 6 cm and PD = 4 cm then find the length of PC.
2. In Fig. 17.19, PA= 4 cm, PB = x + 3, PD = 3 cm and PC = x + 5, find the value of x.
Fig. 17.19 Fig. 17.20 Fig. 17.21
3. If Fig. 17.20, if PA= 4cm, PB = 10 cm, PC = 5 cm, find PD.
4. In Fig. 17.20, if PC = 4 cm, PD = (x + 5) cm, PA = 5 cm and PB = (x + 2) cm,
findx.
5. InFig.17.21,PT= 72 cm,OP=8cm,findtheradiusofthecircle,ifOisthecentre
ofthecircle.
17.7 ANGLES MADE BY A TANGENT AND A CHORD
Let there be a circle with centre O and let XYbe a tangent to the circle at point P. Draw a
chord PQ of the circle through the point P as shown in the Fig. 17.22. Mark a point R on
the major arc PRQ and let S be a point on the minor arc PSQ.
The segment formed by the major arc PRQ and chord PQ is said to be the alternate
segment of ∠QPY and the segment formed by the minor PSQ and chord PQ is said to be
the alternate segment to ∠QPX.
Fig. 17.18

Notes
MODULE - 3
Geometry
Letusseeifthereissomerelationshipbetweenangles
inthealternatesegmentandtheanglebetweentangent
and chord.
Join QR and PR.
Measure ∠PRQ and ∠QPY (See Fig. 17.22)
Whatdoyoufind?Youwillseethat∠PRQ=∠QPY
Repeat this activity with another circle and same or
different radius. You will again find that
∠QPY = ∠PRQ
Now measure ∠QPX and ∠QSP.You will again find that these angles are equal.
Thus, we can state that
The angles formed in the alternate segments by a chord through the point
of contact of a tangent to a circle is equal to the angle between the chord
and the tangent.
Thisresultismorecommonlycalledas“AnglesintheAlternateSegment”.
Let us now check the converse of the above result.
Draw a circle, with centre O, and draw a chord PQ
andletitform∠PRQinalternatesegmentasshown
in Fig. 17.23.
AtP,draw∠QPY=∠QRP.Extendthelinesegment
PY to both sides to form line XY. Join OP and
measure ∠OPY.
What do you observe?You will find that ∠OPY=
90o
showingtherebythatXYisatangenttothecircle.
Repeat this activity by taking different circles and you find the same result. Thus, we can
state that
If a line makes with a chord angles which are equal respectively to the
angles formed by the chord in alternate segments, then the line is a tangent
to the circle.
Let us now take some examples to illustrate:
Example 17.8: In Fig. 17.24, XY is tangent to a
circlewithcentreO.IfAOBisadiameterand∠PAB
= 40o
, find ∠APX and ∠BPY.
Solution: By the Alternate Segment theorem, we
knowthat
Fig. 17.22
Fig. 17.23
Fig. 17.24

Notes
MODULE - 3
Geometry
∠BPY = ∠BAP
∴ ∠BPY = 40o
Again, ∠APB = 90o
(Angleinasemi-circle]
And, ∠BPY + ∠APB + ∠APX = 180o
(Anglesonaline)
∴ ∠APX = 180o
– (∠BPY + ∠APB)
= 180o
– (40o
+ 90o
) = 50o
Example 17.9: In Fig. 17.25, ABC is an isoceles
triangle with AB = AC and XY is a tangent to the
circumcircle of ΔABC. Show that XY is parallel to
base BC.
Solution: In ΔABC, AB =AC
∴ ∠1 = ∠2
Again XYis tangent to the circle atA.
∴ ∠3 = ∠2 (Anglesinthealternatesegment)
∴ ∠1 = ∠3
But these are alternate angles
∴ XY || BC
1. Explainwiththehelpofadiagram,theangleformedbyachordinthealternatesegment
of a circle.
2. In Fig. 17.26, XYis a tangent to the circle with centre O at a point P. If ∠OQP= 40o
,
find the value of a and b.
Fig. 17.26 Fig.17.27
Fig. 17.25

Notes
MODULE - 3
Geometry
3. In Fig. 17.27, PT is a tangent to the circle from an external point P. ChordAB of the
circle, when produced meets TP in P. TA and TB are joined and TM is the angle
bisector of ∠ATB.
If ∠PAT = 40o
and ∠ATB = 60o
, show that PM = PT.
LET US SUM UP
• A line which intersects the circle in two points is called a secant of the circle.
• A line which touches the circle at a point is called a tangent to the circle.
• Atangentisthelimitingpositionofasecantwhenthetwopointsofintersectioncoincide.
• A tangent to a circle is perpendicular to the radius through the point of contact.
• From an external point, two tangents can be drawn to a circle, which are of equal
length.
• IftwochordsABandCDofacircleintersectatapointP(insideoroutsidethecircle),then
PA × PB = PC × PD
• If PAB is a secant to a circle intersecting the circle atAand B, and PTis a tangent to
the circle atT, then
PA × PB = PT2
• Theanglesformedinthealternatesegmentsbyachordthroughthepointofcontactof
a tangent to a circle are equal to the angles between the chord and the tangent.
• Ifalinemakeswithachordangleswhicharerespectivelyequaltotheanglesformed
by the chord in alternate segments, then the line is a tangent to the circle.
TERMINAL EXERCISE
1. Differentitate between a secant and a tangent to a circle with the help of a figure.
2. Showthatatangentisalineperpendiculartothe
radiusthroughthepointofcontact,withthehelp
ofanactivity.
3. In Fig. 17.28, ifAC = BC andAB is a diameter
of circle, find ∠x, ∠y and ∠z.
Fig. 17.28

Notes
MODULE - 3
Geometry
4. In Fig. 17.29, OT = 7 cm and OP = 25 cm, find
the length of PT. If PT′ is another tangent to the
circle, find the length of PT′ and ∠POT.
5. In Fig. 17.30, the perimeter of ΔABC equals
27 cm. If PA= 4 cm, QB = 5 cm, find the length
of QC.
6. In Fig. 17.30, if ∠ABC = 70o
, find ∠BOC.
[Hint: ∠OBC + ∠OCB =
2
1
(∠ABC +
∠ACB)]
7. In Fig. 17.31, AB and CD are two chords of a
circleintersectingattheinteriorpointPofacircle.
If PA= (x + 3) cm, PB = (x – 3) cm, PD = 3 cm
and PC =
3
1
5 cm, find x.
8. In Fig. 17.32, chords BA and DC of the circle,
with centre O, intersect at a point P outside the
circle. If PA= 4 cm and PB = 9 cm, PC = x and
PD = 4x, find the value of x.
9. InFig.17.33,PABisasecantandPTisatangent
tothecirclefromanexternalpoint.IfPT=xcm,
PA= 4 cm andAB = 5 cm, find x.
Fig. 17.29
Fig. 17.33
Fig. 17.32
Fig. 17.31
Fig. 17.30
T’
O

Notes
MODULE - 3
Geometry
10. In Fig. 17.34, O is the centre of the circle and
∠PBQ = 40o
, find
(i) ∠QPY
(ii) ∠POQ
(iii)∠OPQ
ANSWERS TO CHECK YOUR PROGRESS
17.1
1. (i)Perpendicular (ii)equal (iii)pointsofintersection
(iv)two (v)no
2. 50o
, 50o
3. 3 cm
17.2
1. 4.3 cm 2. 3 cm 3. 8 cm
4. 10 cm 4. 6 cm
17.3
2. ∠a = ∠b = 50o
ANSWERS TO TERMINAL EXERCISE
1. ∠x = ∠y = ∠z = 45o
4. PT = 24 cm; PT’ = 24 cm, ∠POT’ = 60o
5. QC = 4.5 6. ∠BOC = 125o
7. x = 5 8. x = 3
9. x = 6
10. (i) 40o
(ii)80o
(iii)50o
Fig. 17.34

SECANTS, TANGENTS AND THEIR PROPERTIES -GEOMETRY

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SECANTS, TANGENTS AND THEIR PROPERTIES -GEOMETRY