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CIRCLE THEOREMS
TANGENTS
A straight line can intersect a circle in three possible ways.
It can be:
A DIAMETER A CHORD A TANGENT
2 points of
intersection
2 points of
intersection
1 point of
intersection
A
B
O O O
A
B
A
TANGENT PROPERTY 1
O
The angle between a
tangent and a radius is a
right angle.
A
TANGENT PROPERTY 2
O
The two tangents drawn
from a point P outside a
circle are equal in length.
AP = BP
A
P
B
O
A
B
P
6 cm
8 cm
AP is a tangent to the circle.
a Calculate the length of OP.
b Calculate the size of angle AOP.
c Calculate the shaded area.
OP2
 62
 82
OP2
 100
OP  10 cm
tanx 
8
6
  
  
 
1 8
tan
6
x
 53.13o
AOP
c Shaded area = area of ΔOAP – area of sector OAB
a b
x

   
     
   
   
2
1 53.13
8 6 6
2 360
 24 16.69
 7.31cm2
(3 s.f.)
Example
CHORDS AND SEGMENTS
major segment
minor segment
A straight line joining two points on the circumference of a
circle is called a chord.
A chord divides a circle into two segments.
SYMMETRY PROPERTIES OF CHORDS 1
O
A B
The perpendicular line from the
centre of a circle to a chord bisects
the chord.
ΙΙ
ΙΙ Note: Triangle AOB is isosceles.
SYMMETRY PROPERTIES OF CHORDS 2
O
A B
If two chords AB and CD are the
same length then they will be the
same perpendicular distance from
the centre of the circle.
ΙΙ
ΙΙ If AB = CD then OP = OQ.
C
D
P
Q
Ι
AB = CD
O
96o
x
Find the value of x.
2x  96  180
2x  84
x  42o
Triangle OAB is isosceles
because OA = OB (radii of circle)
Example
A
B
So angle OBA = x.
THEOREM 1
O
2x
x
The angle at the centre is
twice the angle at the
circumference.
O
96o
x
Find the value of x.
96  2x
x  96  2
x  48o
Angle at centre
= 2 × angle at circumference
Example
O
62o
x
Find the value of x.
x  2  62
x  124o
Angle at centre
= 2 × angle at circumference
Example
O
84o
x
Find the value of x.
84  2x
x  84  2
x  42o
Angle at centre
= 2 × angle at circumference
Example
O
104o
x
Find the value of x.
y  2 104
y  208
Angle at centre
= 2 × angle at circumference
y
x  360  208
x  152o
Example
THEOREM 2
O
An angle in a semi-circle
is always a right angle.
O
Find the value of x.
x  58  90  180
x  32o
Angles in a semi-circle = 90o
and
angles in a triangle add up to 180o
.
58o
x
Example
THEOREM 3
y
x
Opposite angles of a
cyclic quadrilateral add
up to 180o
.
x  y  180o
Find the values of x and y.
x 132  180
x  48o
Opposite angles in a cyclic
quadrilateral add up to 180o
.
x
y
75o 132o
y  75  180
y  105o
Example
THEOREM 4
x
Angles from the same
arc in the same segment
are equal.
x
x
39o
x
Find the value of x.
x  39o
Angles from the same arc in the same
segment are equal.
Example

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circle_theorem_(Extended).ppt

  • 2. TANGENTS A straight line can intersect a circle in three possible ways. It can be: A DIAMETER A CHORD A TANGENT 2 points of intersection 2 points of intersection 1 point of intersection A B O O O A B A
  • 3. TANGENT PROPERTY 1 O The angle between a tangent and a radius is a right angle. A
  • 4. TANGENT PROPERTY 2 O The two tangents drawn from a point P outside a circle are equal in length. AP = BP A P B
  • 5. O A B P 6 cm 8 cm AP is a tangent to the circle. a Calculate the length of OP. b Calculate the size of angle AOP. c Calculate the shaded area. OP2  62  82 OP2  100 OP  10 cm tanx  8 6         1 8 tan 6 x  53.13o AOP c Shaded area = area of ΔOAP – area of sector OAB a b x                    2 1 53.13 8 6 6 2 360  24 16.69  7.31cm2 (3 s.f.) Example
  • 6. CHORDS AND SEGMENTS major segment minor segment A straight line joining two points on the circumference of a circle is called a chord. A chord divides a circle into two segments.
  • 7. SYMMETRY PROPERTIES OF CHORDS 1 O A B The perpendicular line from the centre of a circle to a chord bisects the chord. ΙΙ ΙΙ Note: Triangle AOB is isosceles.
  • 8. SYMMETRY PROPERTIES OF CHORDS 2 O A B If two chords AB and CD are the same length then they will be the same perpendicular distance from the centre of the circle. ΙΙ ΙΙ If AB = CD then OP = OQ. C D P Q Ι AB = CD
  • 9. O 96o x Find the value of x. 2x  96  180 2x  84 x  42o Triangle OAB is isosceles because OA = OB (radii of circle) Example A B So angle OBA = x.
  • 10. THEOREM 1 O 2x x The angle at the centre is twice the angle at the circumference.
  • 11. O 96o x Find the value of x. 96  2x x  96  2 x  48o Angle at centre = 2 × angle at circumference Example
  • 12. O 62o x Find the value of x. x  2  62 x  124o Angle at centre = 2 × angle at circumference Example
  • 13. O 84o x Find the value of x. 84  2x x  84  2 x  42o Angle at centre = 2 × angle at circumference Example
  • 14. O 104o x Find the value of x. y  2 104 y  208 Angle at centre = 2 × angle at circumference y x  360  208 x  152o Example
  • 15. THEOREM 2 O An angle in a semi-circle is always a right angle.
  • 16. O Find the value of x. x  58  90  180 x  32o Angles in a semi-circle = 90o and angles in a triangle add up to 180o . 58o x Example
  • 17. THEOREM 3 y x Opposite angles of a cyclic quadrilateral add up to 180o . x  y  180o
  • 18. Find the values of x and y. x 132  180 x  48o Opposite angles in a cyclic quadrilateral add up to 180o . x y 75o 132o y  75  180 y  105o Example
  • 19. THEOREM 4 x Angles from the same arc in the same segment are equal. x x
  • 20. 39o x Find the value of x. x  39o Angles from the same arc in the same segment are equal. Example