1. Lessons 30 -35
Information
on
Circles

2. Lesson 30 & 31 Notes
Terminology
Circle with centre k diameter
semicircles
chord
Radius from centre to outside edge
Perpendicular bisector central angle
Inscribed angle tangent
The properties that should be discovered and used are as follows:
k
point of tangency

3. EC = 3.19 cm
FE = 3.19 cm
m∠AEC = 90.00°
E
A
F
C
A) The perpendicular (line) from the centre of a circle to a chord bisects the chord
This also works in the reverse. If you find
the middle point of any chord and draw a
perpendicular line from the middle point it
will pass through the center of the circle.
This property can help you find the center
because, if you apply this property more than
once inside one circle, the point where the
perpendicular lines intersect is where the
center of the circle is.
B) The measure of the central angle is equal to twice the measure of the inscribed angle
subtended (shared) by the same arc.
This property becomes very useful for
finding the center when we start to
create central angles of 1800
. This means
that the inscribed angle will be 900
.
Center
F
This
angle
will be
half
the
central
This Angle will
be twice as much
as the inscribed
The Arc that is
shared is called
the Subtended
Arc

4. If students create a 900
angle on the circumference of the circle the end of each ray
will be the diameter. If you apply this property more than once, the point where
the diameters insect will be the center of the
circle.
C) The inscribed angles subtended by the same arc are congruent.
Again, this is a property that only works
when the inscribed angles share an arc.
If you are using computer software, ask
the students to drag one of the vertices
into the subtended arc and see what
happens to the measurements. They
will no longer be congruent because
they are no longer sharing the same arc.
D) A tangent to a circle is perpendicular to the radius at the point of
tangency.
This becomes a useful tool for
finding the center of a circle. If we
extend the radius to a diameter, and
apply the property more than once,
the intersection point will be the
center.
m∠JML = 42.94°
m∠JKL = 42.94°
L
J
K
M

5. I
H
G
J
E) An angle inscribed in a semi-circle
is ALWAYS 900
.
This is directly related item B in the
list. Do this twice and the point where
the hypotenuses intersect will be the
center of the circle.
F) The opposite angle of a cyclic
quadrilateral (are quadrilaterals inside a circle with its vertices on the
circumference of the circle) are supplementary (add up to 1800
).
G) An exterior angle of a cyclic quadrilateral is equal to the interior opposite
angle.
40HJG∠ = °and 110HGK∠ = °
Then we know that angle JGH =
700
because it is in a straight line
with HGK (180 – 110 = 70)
From this we can then figure out
that 70GHJ∠ = ° because interior
angles of 180GHJ∆ = °
If we know that 70GHJ∠ = ° we also
know that 70GIJ∠ = ° because both are
inscribed angles subtended by the
minor arc JG.
180I G+ = °R R
K
I
H
G
J

6. O
E
D
C F
Examples
1) Determine the measures of angles
ECB
BDC
BAD
DBE
∠
∠
∠
∠
where E is the center of the circle.
2) Determine the length of a chord AC.
3) Given a circle with the center O, and OF CD⊥ and DE=20 and OF=DF find the
length of OF
EO
DF
CD
30
100
E
C
A
B
D
610
B
A
C

7. O
A BC
Lesson 32 – Properties of Tangents
1) Tangent segments to a circle’s circumference from any
external point are congruent in length.
2) The angle between the tangent and the chord is
half the measure of the intercepted arc.
3) The angle between the tangent and the chord is equal to the inscribed angle on
the opposite side of the chord.
Examples
1) The circle has a radius of 5 and AB is
tangent to the circle at point C.
50B∠ = ° and OA = OB
Find the Area of AOB∆

8. 2)
Find the length of AC
3) Given that ED is tangent at c
60
(2 )
(3 5)
BCA
ACD x
BCE x
Find B
∠ = °
∠ = °
∠ = − °
∠
5
30
50
A
D
B
C
60

9. 4) 15
AD BC
ACB
Find DBC
∠ = °
∠
P
Lesson 33 – No Notes – Just Review
Lesson 34 – Circle and Polygon Properties
Example 1 – If diameter CD is perpendicular to a chord AB at E, verify that
ABC∆ is isosceles.
This means we want to prove AC=CB

10. 8
6
4
2
-2
-4
-6
-8
-10
-15 -10 -5 5 10 15
Example 2)
Determine the measure of BAC∠ if 60 70DEF and EFC∠ = ° ∠ = °
Example 3)
Find the center of the circle that passes through the points (0,0), (0,6) and (4,0).

11. Example 4)
With a circle centered at (0,0) and tangent to the circle at (3,4) find the equation of
AB
10
8
6
4
2
-2
-4
-6
-8
-15 -10 -5 5 10 15