A presenatation on Circle containing all the theoram and HOTs

2. .
.
O
B
Radius
Q
P
X
Chord
Centre
of the
Circle
In the above circle, OB is the radius and PQ is any
chord. OX is drawn perpendicularly to PQ which
meets it at X. OX bisects PQ and OXQ is 90.

3. •A line segment whose end points lie on the circle is
called a chord to the circle.
B
A Here, AB is the
chord.
•If a circle and a line have no common point, then
the line is called a non-intersecting line with respect
to the circle.
P
Q
Here, PQ is the non-intersecting
line.

4. •If a circle and a line have two common points or a line
intersects a circle in two distinct points, then the line is
called a secant to the circle.
P
Q
A
B
•If a line and a circle have only one point in common,
or a line intersects a circle at only one point, then it is
called a tangent to the circle.
P
Q
A
Here AB is the
secant.
Here, PQ is the tangent to the
circle. PQ meets the circle at A.

5. TANGENT
A Line meeting a circle only in one one point is
called a tangent to the circle at that point.
The point at which the tangent line meets the
circle is called the point of contact. Here A is
the point of contact.

6. NUMBER OF TANGENTS
1. There is one and only one tangent passing
through a point lying on a circle.
2. There are exactly two tangents from a
point lying outside a circle.
3. We can draw infinitely many tangents to a
circle as there are infinitely many points
on the circumference of a circle.

7. •There is only one tangent at a point on the
circumference of the circle.
•The common point of the tangent and the circle is
called the point of contact.
•The tangent at any point of a circle is
perpendicular to the radius through the point of
contact.
•There is no tangent to the circle passing through a
point lying inside a circle.
Q
P
A
O.

8. To prove that OT is perpendicular to AB
•Assume that OT is not perpendicular to AB
Then there must be a point, D say, on AB such
that OD is perpendicular to AB.
D
C
•Since ODT is a right angle then angle OTD is
acute (angle sum of a triangle).
•But the greater angle is opposite the greater
side therefore OT is greater than OD.
•But OT = OC (radii of the same circle)
therefore OC is also greater than OD, the
part greater than the whole which is
impossible.
•Therefore OD is not perpendicular to AB.
•By a similar argument neither is any other
straight line except OT.
•Therefore OT is perpendicular to AB.
To prove that the angle between a tangent and a radius drawn to the
point of contact is a right angle.
O
A
B
T

12. To prove that the angle between a tangent and a chord through the
point of contact is equal to the angle subtended by the chord in the
alternate segment.
A
T
B
C
D
O
To prove that angle BTD = angle TCD
•With centre of circle O, draw straight lines
OD and OT.
•Let angle DTB be denoted by .

•Then angle DTO = 90 -  (Theorem 4 tan/rad)
90 - 
•Also angle TDO = 90 -  (Isos triangle)
90 - 
•Therefore angle TOD = 180 –(90 -  + 90 - )
= 2 (angle sum triangle)
2
•Angle TCD =  (Theorem 1 angle at the centre)

•Angle BTD = angle TCD

13. Draw a circle and two lines parallel to a given line such that one is a
tangent and the other, a secant to the circle.
It can be observed that AB and
CD are two parallel lines. Line
AB is intersecting the circle at
exactly two points, P and Q.
Therefore, line AB is the secant
of this circle. Since line CD is
intersecting the circle at exactly
one point, R, line CD is the
tangent to the circle

14. Prove that the tangents drawn at the ends of a diameter of a circle are
parallel.
Let AB be a diameter of the circle. Two
tangents PQ and RS are drawn at points A and
B respectively.
Radius drawn to these tangents will be
perpendicular to the tangents.
Thus, OA ⊥ RS and OB ⊥ PQ
∠OAR = 90º
∠OAS = 90º
∠OBP = 90º
∠OBQ = 90º
It can be observed that
∠OAR = ∠OBQ (Alternate interior angles)
∠OAS = ∠OBP (Alternate interior angles)
Since alternate interior angles are equal, lines
PQ and RS will be parallel.