8 8.1tangents to a circle 8.2tangents to a circle from an external point chapter summary case study...
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8
8.1 Tangents to a Circle
8.2 Tangents to a Circle from an External Point
Chapter Summary
Case Study
8.3 Angles in the Alternate Segments
Basic Properties of Circles (2)
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Case Study
The wheels of a train and the rails illustrate an important geometrical relationship between circles and straight lines.
When the train travels on the rails, it shows how a circle and a straight line touch each other at only one point.
Can you give me onereal-life example of a circle and a straight line?
Yes, the wheel of a train is a circle and the rail is a straight line.
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8.1 Tangents to a Circle8.1 Tangents to a Circle
We can draw a straight line AB and a circle in three different ways:Case 1: The straight line does not meet the circle.
Case 2: The straight line cuts the circle at two distinct points, P and Q.
Case 3: The straight line touches the circle at exactly one point, T.
In case 3, at each point on a circle, we can draw exactly one straight line such that the line touches the circle at exactly one point.
Tangent to a circle:straight line if and only if touching the circle at exactly one point
Point of contact (point of tangency):point common to both the circle and the straight line
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8.1 Tangents to a Circle8.1 Tangents to a Circle
There is a close relationship between the tangent to a circle and the radius joining the point of contact:
Theorem 8.1If AB is a tangent to the circle with centre O atT, then AB is perpendicular to the radius OT.Symbolically, AB OT. (Reference: tangent radius)
This theorem can be proved by contradiction:
Suppose AB is not perpendicular to the radius OT.Then we can find another point T on AB such that OT AB.Using Pythagoras’ Theorem, OT is shorter than OT. Thus T lies inside the circle.
∴ AB cuts the circle at more than one point.
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8.1 Tangents to a Circle8.1 Tangents to a Circle
The converse of Theorem 8.1 is also true:
Theorem 8.2OT is a radius of the circle with centre O andAB is a straight line that intersects the circleat T. If AB is perpendicular to OT, then ABis a tangent to the circle at T.In other words, if AB OT,
then AB is a tangent to the circle at T. (Reference: converse of tangent radius)
The perpendicular to a tangent at its pointof contact passes through the centre ofthe circle.
Hence we can deduce an important fact:
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8.1 Tangents to a Circle8.1 Tangents to a Circle
Example 8.1T
Solution:
In the figure, O is the centre of the circle. AB is a tangent to the circle at T. OC TC 9 cm. (a) Find CAT and CTA. (b) Find the length of AT.
(a) OT OC 9 cm (radii)
∴ OCT is an equilateral triangle. ∴ COT OTC 60 (prop. of equilateral )
OTA 90 (tangent radius)
In OAT,CAT OTA COT 180 ( sum of )
∴ CTA 90 60 30
CAT 30
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P. 7
8.1 Tangents to a Circle8.1 Tangents to a Circle
Example 8.1T
Solution:
In the figure, O is the centre of the circle. AB is a tangent to the circle at T. OC TC 9 cm. (a) Find CAT and CTA. (b) Find the length of AT.
(b) ∵ CTA CAT 30 (proved in (a))
∴ CA CT 9 cm (sides opp. equal s)
In OAT,AT 2 OT 2 OA 2 (Pyth. Theorem)
AT cm22 918 cm39
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P. 8
8.1 Tangents to a Circle8.1 Tangents to a Circle
Example 8.2T
Solution:
In the figure, AB is a tangent to the circle at T. POQB is a straight line. If ATP 65, find TBQ.
OTA 90 (tangent radius)
∴ OTP 90 65 25
Join OT.
∵ OP OT (radii)
∴ OPT OTP 25 (base s, isos. )In BPT,ATP TBQ OPT (ext. of )
65 TBQ 25TBQ 40
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P. 9
8.2 Tangents to a Circle from8.2 Tangents to a Circle from an External Point an External Point
Consider an external point T of a circle.We can always draw two tangents from that point.
In the figure, we can prove that OTA OTB (RHS):
OAT OBT 90 (tangent radius)
OT OT (common side)
OA OB (radii)
Hence the corresponding sides and the corresponding angles of OTA and OTB are equal:
TA TB (corr. sides, s)
TOA TOB (corr. s, s)
OTA OTB (corr. s, s)
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P. 10
8.2 Tangents to a Circle from8.2 Tangents to a Circle from an External Point an External Point
Properties of tangents from an external point:
Theorem 8.3In the figure, if TA and TB are the two tangentsdrawn to the circle with centre O froman external point T, then(a) the lengths of the two tangents are
equal, that is, TA TB; (b) the two tangents subtend equal angles at the centre,
that is, TOA TOB; (c) the line joining the external point to the centre of the
circle is the angle bisector of the angle included by the two tangents, that is, OTA OTB.
(Reference: tangent properties)
In the figure, OT is the axis of symmetry.
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P. 11
8.2 Tangents to a Circle from8.2 Tangents to a Circle from an External Point an External Point
Example 8.3T
Solution:
In the figure, TA and TB are tangents to the circle with centre O. If ABT 65, find(a) ATB, (b) AOB.
(a) ∵ TA TB (tangent properties)
∴ TAB TBA (base s, isos. )
65In TAB,ATB 2(65) 180 ( sum of )
ATB 50
(b) OAT OBT 90 (tangent radius)
∴ AOB OAT ATB OBT 360 ( sum of polygon)
AOB 90 50 90 360 AOB 130
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P. 12
8.2 Tangents to a Circle from8.2 Tangents to a Circle from an External Point an External Point
Example 8.4T
Solution:
In the figure, TA and TC are tangents to the circle with centre O.
If AB : BC 1 : 2 and ADC 66, find x and y.
( (
ABC 66 180 (opp. s, cyclic quad.)
ABC 114ACB : BAC AB : BC (arcs prop. to s at ⊙ce)
( (
x : BAC 1 : 2∴ BAC 2x
In ABC,ABC BAC x 180 ( sum of )
114 2x x 180 x 22
2x
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P. 13
8.2 Tangents to a Circle from8.2 Tangents to a Circle from an External Point an External Point
Example 8.4T
Solution:
In the figure, TA and TC are tangents to the circle with centre O.
If AB : BC 1 : 2 and ADC 66, find x and y.
( (
AOC 2 66 ( at the centre twice at ⊙ce)
OAT OCT 90 (tangent radius)
132
∴ AOC OAT ATC OCT 360 ( sum of polygon)
132 90 ATC 90 360 ATC 48
∵ TC TA (tangent properties)
∴ TCA TAC (base s, isos. )
In TAC,ATC 2TAC 180 ( sum of )
TAC 66
∵ BAC 2x 44
∴ y 22
2x
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P. 14
8.2 Tangents to a Circle from8.2 Tangents to a Circle from an External Point an External Point
Example 8.5T
Solution:
The figure shows an inscribed circle in a quadrilateral ABCD.If AB 16 cm and CD 12 cm, find the perimeter of the quadrilateral.
Referring to the figure,AP AS, BP BQ, CQ CRand DR DS. (tangent properties)
S
P
Q
R
Let AP AS a, BP BQ b, CQ CR c and DR DS d.Then a b 16 cm and c d 12 cm.∵ DA AS SD
a dand BC BQ QC
b c ∴ Perimeter 16 cm (b c) 12 cm (a d)
16 cm 12 cm a b c d 56 cm
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P. 15
8.3 Angles in the Alternate Segments8.3 Angles in the Alternate Segments
In the figure, AB is a tangent to the circle at T and PT is a chord of the circle.
Tangent-chord angles:angles formed between a chord and a tangent to a circle, such as PTA and PTB.
Alternate segment:segment lying on the opposite side of a tangent-chord angle
segment I is the alternate segment with respect to PTB segment II is the alternate segment with respect to PTA
Consider the tangent-chord angle b.Then a is an angle in the alternate segment with respect to b.
Notes:We can construct infinity many angles in the alternate segment with respect to b.
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P. 16
8.3 Angles in the Alternate Segments8.3 Angles in the Alternate Segments
The figure shows another angle in the alternate segment with respect to b with BR passing through the centre O.
R C a ( in the same segment)RAB 90 ( in semicircle)
In ABR,R RAB ABR 180 ( sum of )
ABR 90 a∵ ABR ABQ 90 (tangent radius)∴ (90 a) b 90
a b
a 90 ABR 180
Theorem 8.4A tangent-chord angle of a circle is equalto any angle in the alternate segment.Symbolically, a b and p q.(Reference: in alt. segment)
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P. 17
8.3 Angles in the Alternate Segments8.3 Angles in the Alternate Segments
Example 8.6T
Solution:
In the figure, TS is a tangent to the circle. TBC is a straight line. BA BT and ATB 48. (a) Find ACB. (b) Find CAS.
(a) ∵ BA BT (given)
∴ BAT BTA (base s, isos. )
48 ∴ ACB BAT ( in alt. segment)
48
(b) CBA BTA BAT (ext. of )
96 ∴ CAS CBA ( in alt. segment)
96
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P. 18
8.3 Angles in the Alternate Segments8.3 Angles in the Alternate Segments
Example 8.7T
Solution:
The figure shows an inscribed circle of ABC. The circle touches the sides of the triangle at P, Q and R respectively. If BAC 40 and ACB 68, find all the angles in PQR.
∵ AP AR (tangent properties)
∴ APR ARP (base s, isos. )
In PAR,40 APR ARP 180 ( sum of )
ARP 70
Similarly, ∵ CQ CR (tangent properties)
∴ CRQ CQR 56
∴ PQR ARP 70 ( in alt. segment)
∴ QPR CRQ 56 ( in alt. segment)
∴ PRQ 180 70 56 54
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P. 19
8.3 Angles in the Alternate Segments8.3 Angles in the Alternate Segments
The converse of Theorem 8.4 is also true:
Theorem 8.5A straight line is drawn through an endpoint of a chord of a circle. If the anglebetween the straight line and the chordis equal to an angle in the alternatesegment, then the straight line is atangent to the circle. In other words, if x y, then TA is atangent to the circle at A. (Reference: converse of in alt. segment)
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P. 20
8.3 Angles in the Alternate Segments8.3 Angles in the Alternate Segments
Example 8.8T
Solution:
In the figure, AB // PQ and CD is a common chord of the circles. Prove that PQ is a tangent to the larger circle.
BAC CQP (alt. s, AB // PQ)
BAC CDQ (ext. , cyclic quad.)
∴ CQP CDQ
∴ PQ is a tangent to the larger circle. (converse of in alt. segment)
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P. 21
8.1 Tangents to a Circle
Chapter Summary
1. If AB is a tangent to the circle with centre O at T, then AB is perpendicular to the radius OT.
(Ref: tangent radius)
2. OT is a radius of the circle with centre O and ATB is a straight line. If AB is perpendicular to OT, then AB is a tangent to the circle at T.
(Ref: converse of tangent radius)
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P. 22
Chapter Summary8.2 Tangents to a Circle from an External Point
If TA and TB are tangents to the circle with centre O,from an external point T, then(a) TA TB;
(The length of the two tangents are equal.)(b) TOA TOB;
(Two tangents subtend equal angles at the centre.)(c) OTA OTB.
(OT bisects the angle between the two tangents.)
(Ref: tangent properties)
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P. 23
Chapter Summary8.3 Angles in the Alternate Segments
1. If TA is a tangent to the circle, then x y and p q.
(Ref: in alt. segment)
2. If x y, then TA is a tangent.(Ref: converse of in alt. segment)