Transcript Book 4 Chapter 8 Basic Properties of Circles (2)

8
Basic Properties of
Circles (2)
Case Study
8.1
Tangents to a Circle
8.2
Tangents to a Circle from an External Point
8.3
Angles in the Alternate Segments
Chapter Summary
Case Study
Can you give me one
real-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.
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.
P. 2
8.1 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
P. 3
8.1 Tangents to a Circle
There is a close relationship between the tangent to a circle and
the radius joining the point of contact:
Theorem 8.1
If AB is a tangent to the circle with centre O at
T, 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.
P. 4
8.1 Tangents to a Circle
The converse of Theorem 8.1 is also true:
Theorem 8.2
OT is a radius of the circle with centre O and
AB is a straight line that intersects the circle
at T. If AB is perpendicular to OT, then AB
is 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)
Hence we can deduce an important fact:
The perpendicular to a tangent at its point
of contact passes through the centre of
the circle.
P. 5
8.1 Tangents to a Circle
Example 8.1T
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.
Solution:
(a) OT  OC  9 cm
∴ DOCT is an equilateral triangle.
∴ COT  OTC  60
OTA  90
∴ CTA  90  60
 30
In DOAT,
CAT  OTA  COT  180
CAT  30
(radii)
(prop. of equilateral D)
(tangent  radius)
( sum of D)
P. 6
8.1 Tangents to a Circle
Example 8.1T
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.
Solution:
(b) ∵ CTA  CAT  30
∴ CA  CT  9 cm
In DOAT,
AT 2  OT 2  OA 2
AT  182  92 cm
 9 3 cm
(proved in (a))
(sides opp. equal s)
(Pyth. Theorem)
P. 7
8.1 Tangents to a Circle
Example 8.2T
In the figure, AB is a tangent to the circle at T. POQB is a
straight line. If ATP  65, find TBQ.
Solution:
Join OT.
OTA  90
∴ OTP  90  65
 25
∵ OP  OT
∴ OPT  OTP  25
In DBPT,
ATP  TBQ  OPT
65  TBQ  25
TBQ  40
(tangent  radius)
(radii)
(base s, isos. D)
(ext.  of D)
P. 8
8.2 Tangents to a Circle from
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 DOTA  DOTB (RHS):
OAT  OBT  90
(tangent  radius)
OT  OT
(common side)
OA  OB
(radii)
Hence the corresponding sides and the corresponding angles of
DOTA and DOTB are equal:
TA  TB
(corr. sides,  Ds)
TOA  TOB
(corr. s,  Ds)
OTA  OTB
(corr. s,  Ds)
P. 9
8.2 Tangents to a Circle from
an External Point
Properties of tangents from an external point:
Theorem 8.3
In the figure, if TA and TB are the two tangents
drawn to the circle with centre O from
an 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.
P. 10
8.2 Tangents to a Circle from
an External Point
Example 8.3T
In the figure, TA and TB are tangents to the circle with centre O.
If ABT  65, find
(a) ATB,
(b) AOB.
Solution:
(a) ∵ TA  TB
∴ TAB  TBA
 65
In DTAB,
ATB  2(65)  180
ATB  50
(tangent properties)
(base s, isos. D)
( sum of D)
(b) OAT  OBT  90
∴ AOB  OAT  ATB  OBT  360
AOB  90  50  90  360
AOB  130
(tangent  radius)
( sum of polygon)
P. 11
8.2 Tangents to a Circle from
an External Point
Example 8.4T
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.
2x
Solution:
(
(
ABC  66  180 (opp. s, cyclic quad.)
ABC  114
ACB : BAC  AB : BC
(arcs prop. to s at ⊙ce)
x : BAC  1 : 2
∴
BAC  2x
In DABC,
ABC  BAC  x  180 ( sum of D)
114  2x  x  180
x  22
P. 12
8.2 Tangents to a Circle from
an External Point
Example 8.4T
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.
2x
Solution:
AOC  2  66 ( at the centre twice  at ⊙ce)
 132
OAT  OCT  90
(tangent  radius)
∴ AOC  OAT  ATC  OCT  360
132  90  ATC  90  360
ATC  48
∵ TC  TA
(tangent properties)
∴ TCA  TAC
(base s, isos. D)
In DTAC,
ATC  2TAC  180
( sum of D)
TAC  66
( sum of polygon)
∵
∴
BAC  2x
 44
y  22
P. 13
8.2 Tangents to a Circle from
an External Point
Example 8.5T
The figure shows an inscribed circle in a quadrilateral ABCD.
If AB  16 cm and CD  12 cm, find the perimeter of the quadrilateral.
S
Solution:
P
Referring to the figure,
AP  AS, BP  BQ, CQ  CR
and DR  DS.
R
Q
(tangent properties)
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
and BC  BQ  QC
ad
bc
∴ Perimeter  16 cm  (b  c)  12 cm  (a  d)
 16 cm  12 cm  a  b  c  d
 56 cm
P. 14
8.3 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.
P. 15
8.3 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 DABR,
R  RAB  ABR  180 ( sum of D)
a  90  ABR  180
ABR  90  a
∵
ABR  ABQ  90
(tangent  radius)
∴
(90  a)  b  90
ab
Theorem 8.4
A tangent-chord angle of a circle is equal
to any angle in the alternate segment.
Symbolically, a  b and p  q.
(Reference:  in alt. segment)
P. 16
8.3 Angles in the Alternate Segments
Example 8.6T
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.
Solution:
(a) ∵ BA  BT
∴ BAT  BTA
 48
∴ ACB  BAT
 48
(given)
(base s, isos. D)
( in alt. segment)
(b) CBA  BTA  BAT
(ext.  of D)
 96
∴ CAS  CBA ( in alt. segment)
 96
P. 17
8.3 Angles in the Alternate Segments
Example 8.7T
The figure shows an inscribed circle of DABC. 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 DPQR.
Solution:
∵ AP  AR
(tangent properties)
∴ APR  ARP
(base s, isos. D)
In DPAR,
40  APR  ARP  180 ( sum of D)
ARP  70
∴ PQR  ARP  70 ( in alt. segment)
Similarly,
∵ CQ  CR
(tangent properties)
∴ CRQ  CQR  56
∴ QPR  CRQ  56 ( in alt. segment)
∴
PRQ
 180  70  56
 54
P. 18
8.3 Angles in the Alternate Segments
The converse of Theorem 8.4 is also true:
Theorem 8.5
A straight line is drawn through an end
point of a chord of a circle. If the angle
between the straight line and the chord
is equal to an angle in the alternate
segment, then the straight line is a
tangent to the circle.
In other words, if x  y, then TA is a
tangent to the circle at A.
(Reference: converse of  in alt. segment)
P. 19
8.3 Angles in the Alternate Segments
Example 8.8T
In the figure, AB // PQ and CD is a common chord of the circles.
Prove that PQ is a tangent to the larger circle.
Solution:
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)
P. 20
Chapter Summary
8.1 Tangents to a Circle
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)
P. 21
Chapter Summary
8.2 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)
P. 22
Chapter Summary
8.3 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)
P. 23