Sector:
A sector of a circle is a pie shaped portion of the circle area, and it
is between two segments coming out of the center of the circle (Fig: 16).
Another way to say, it is the region enclosed by the central angle of
a circle and the circle itself. A segment of a circle is the region
enclosed by a chord and the arc that the chord defines (Fig: 17).
The figure given below will help you understand it easily:
For example:
Find the area of the segment
of a circle in Fig: 17,
(Continued from above)
Given: the central angle of the segment is 60
and the radius is 8cm.
Step 1: Area of sector = (1 / 6)*(r2) = (1 / 6)*(*
82)
= (1 / 6) * [ * (64)] = 33.49 cm2
Step 2: Area of triangle OAB = (823)
/ 4 = 27.71 cm2
Step 3: Area of segment = (33.49 – 27.71)
= 5.78 cm2
Area of segment is 5.78 cm2, as the final answer.
Tangents:
A tangent to a circle is a line in plane of the circle that intersects
the circle in one and only one point (Fig: 18).
Secant
A secant of a circle is a line that intersects the circle in two points.
Postulate
At a given point on a given circle, one and only one line can be drawn
that is tangent to the circle.
Theorems:
• if a line is perpendicular to a radius at a point on the circle, then the
line is tangent to the circle.
• if a line is tangent to a circle, then it is perpendicular to the radius at a point
on the circle.
• a line is tangent to a circle if and only if it is perpendicular to a radius at
its point of intersection with the circle.
Common Tangents
A common tangent is a line that is tangent to each of two circles.
Chords, Secants, Tangents in Circles: Rules to work:
Segments Formed by Two Intersecting Chords
Theorem 1:
If two chords intersect within a circle, the product of the measures of the segments of one chord is equal to the product of the measures of the segments of the other.
Given Chords AB and CD intersect at E in the interior of circle
O.
Prove (AE)*(EB) = (CE) (ED)
Proof: Draw AD and CB.
Statements
Reasons
1
Chord
1
Given
2
Draw
2
Two points determine only one line.
3
AC
and
DB
3
Inscribed angles of a circle that intercept the same arc are congruent
4
ADE
~ CBE
4
By AA~
5
AE / CE = ED / EB
5
The lengths of the corresponding sides of similar triangles are in proportion.
6
(AE )*(EB ) = (CE )*(ED)
6
In a proportion, the product of the means equals the product of the extremes.
Segments Formed by a Tangent Intersecting a Secant
External Segment: The part of the secant segment that is outside
the circle, is called the external segment of the secant.
Theorem 2:
If a tangent and a secant are drawn to a circle from an external point, then the square of the length of the tangent segment is equal to the
product of the lengths of the secant segment and its external segment i.e. (PA)2 = (PC)* (PB)
Segments Formed by Intersecting Secants:
Theorem 3:
If two secant segments are drawn to a circle from an external point;
then product of the length of one secant segment and its external segment,
equals the product of the length of the other secant segment and its
external segment i.e.(AC)*(AB)
= (AE)*(AD)
For example, look at the word problem where you can use the above basic properties of the
segments formed by secants and tangents and also solve basic linear and/or quadratic
equations.
Given: in circle O below (NTS), two secants from point P intersect circle
O such that arcs CP = 10, BP = 9, CA = 2 x and
BD = 2x +3. Find out the measure of
segment AP?
You know from the above theorem: "the products of the external segment and the entire
secant must be equal for both secants". Thus,
Step 1:
CP (CP + CA) = BP (BP + BD)
10(10 + 2x) = 9(2x + 3+ 9)
Step 2: Solve the equation for x we get,
100 + 20x = 18x + 108
2x = 8
x = 4
Step 3: As AP equals (2x + 10)
AP = [2(4) + 10]
AP = 18 units, as the final answer.
Angles involving tangents and secants:
When two secants intersect inside a circle, the measure of each angle formed
is related to one-half the sum of the measures of the intercepted arcs.
The figure below explains it:
For example, in the figure above (NTS: Not to Scale),
arc and
arc
are 65 and
55, respectively. In line with the
above theorem, the measures of both
1
and 2
in the figure are 60.
When the secants intersect outside of circles, the measure of the angle formed is equal to
one-half the difference of the degree measures of the intercepted arcs.
E.g., Given:
arc
= 65
and arc
= 105.
Find the m1.
By the theorem stated above, the measure of 1
= 1/2 (
–
)
i.e. 1
= 1/2 ( 105
– 65)
= 20
, as the final answer.
Remember: Chords, Secants, and Tangents in Circles - Rules to work,
•
Intersecting Chords Rule: (segment piece) * (segment piece) = (segment piece)
* (segment piece).
•
Secant-Tangent Rule: (whole secant) * (external part) = (tangent)2
•
Secant-Secant Rule: (whole secant) * (external part) = (whole secant) * (external
part)
•
A chord is a straight line joining two points on the circumference. The longest chord is the
diameter and diameter passes through the center. E.g. A and
B are the points on the circle.
When we join these two points, AB
is the chord.
•
A sector is a region enclosed by two radii and an arc. In a circle with center at O and A, B two points on the circumference, AOB
is the angle subtended by the
arc AB at the centre O. The larger region is called the major sector and the smaller
one, a minor sector.
•
A segment of a circle is the region enclosed by a chord and an arc of the circle.
The larger segment is the major segment and the smaller one, the minor segment. •
A secant is a line intersecting the circle at two distinct points. If point A
and B
are outside the circle and we join these, it intersects circle at two distinct points. •
If a line and a circle have only one intersection point, this line is called a tangent.
It is always perpendicular to the radius drawn to the point of intersection.
This property is abbreviated as tan
radius.
•
The diameter of a circle divides the circle into equal halves and each part is called a semi circle.
The video above will explain more in detail with the help of several examples;
about the chords, tangents, secants, sector and segments,
which will give you
deeper understanding and provide Geometry help.
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and the radius is 8cm.
r2) = (1 / 6)*(
3)
/ 4 = 27.71 cm2 
A
ADE
~ 
and
arc 
are 65
= 65
= 105
radius.