CIRCLE
Circle:
A circle is
a shape consisting of all points in a plane that are a given distance from a
given point, the centre;
equivalently it is the curve traced out by a point that moves in a plane so
that its distance from a given point is constant.
The distance between any point of the circle and the centre is called the radius.
Center of a Circle
:
The
center of a circle is the center point in a circle from which all the distances
to the points on the circle are equal. This distance is called the radius of
the circle.
Here,
point P is the center of the circle.
Semicircle:
A
semi-circle is half of a circle, formed by cutting a whole circle along a line
segment passing through the center of the circle. This line segment is called
the diameter of the circle.
Quarter Circle:
A
quarter circle is a quarter of a circle, formed by splitting a circle into 4
equal parts or a semicircle into 2 equal parts.
A
quarter circle is also called a quadrant.
Tangent to a
Circle
The line that joins two infinitely close points from a
point on the circle is a Tangent. In other words, we can say that
the lines that intersect the circles exactly in one single point are Tangents.
Point of tangency is the point where the tangent touches the circle. At the
point of tangency, a tangent is perpendicular to the radius. Several theorems
are related to this because it plays a significant role in geometrical constructions and proofs. We will look at them one by
one.
Tangents
Formula
The formula for the tangent is given below. For the
description of formula, please look at the following diagram.
Here, we consider a circle where P is the exterior point.
From that exterior point, the circle has the tangent at a points A and B. A
straight line which cuts curve into two or more parts is known as a secant. So,
here secant is PR is drawn and at Q, R intersects the circle as shown in the
upper diagram. The formula for tangent-secant states that:
PR/PS = PS/PQ
PS2 =
PQ.PR
Properties of
Tangents
Remember the following points about the properties of
tangents-
The tangent line
never crosses the circle, it just touches the circle.
At the point of
tangency, it is perpendicular to the radius.
A chord and
tangent form an angle and this angle is same as that of tangent inscribed on
the opposite side of the chord.
From the same external
point, the tangent segments to a circle are equal.
Theorems for
Tangents to Circle
Theorem 1
A radius is obtained by joining the centre and the point
of tangency. The tangent at a point on a circle is at right angles to this
radius. Just follow this below diagram: Here AB⊥OP
Theorem 2
This theorem states that if from one external point, two
tangents are drawn to a circle then they have equal tangent segments. Tangent
segment means line joining to the external point and the point of tangency. Consider
the following diagram: Here, AC=BC.
These are some of the basic
theorems on tangents to a circle. In the next section, we will look at some of
the basic properties of tangents.
Construction of a tangent
to a circle (Using the centre)
Draw a circle of radius 3 cm.
Take a point P on this circle and draw a tangent
at P.
Solution
Given,
radius r = 3 cm
Construction
Step 1: Draw a circle with centre at O of
radius 3 cm.
Step 2: Take
a point P on the circle. Join OP.
Step 3: Draw perpendicular line
TT’ to OP which passes through P.
Step 4: TT ‘ is the
required tangent.
Construct of a tangent to
a circle (Using alternate segment theorem)
Draw a circle of radius 4 cm.
At a point L on it draw a tangent to the circle using
the alternate segment.
Solution
Given,
radius=4 cm
Construction
Step 1 :
With O as the centre, draw a circle of radius 4 cm.
Step 2: Take a point L on
the circle. Through L draw any chord LM.
Step 3: Take a point M distinct
from L and N on the circle, so that L, M and N are in anti-clockwise direction. Join LN and NM.
Step 4: Through L draw
a tangent TT ’ such that ∠TLM = ∠MNL.
Step 5: TT ‘ is the required tangent.
Construction of pair of
tangents to a circle from an external point P.
Draw a circle of diameter 6 cm from
a point P, which is 8 cm away from
its centre. Draw the two tangents PA and PB to
the circle and measure their lengths.
Solution
Given,
diameter (d) = 6 cm, we find radius (r) = 6/2 = 3 cm
Construction
Step 1: With centre at O,
draw a circle of radius 3 cm.
Step 2: Draw a line OP of
length 8 cm.
Step 3: Draw a perpendicular bisector
of OP, which cuts OP at M.
Step 4: With M as
centre and MO as radius, draw a circle which cuts previous
circle at A and B.
Step5: Join AP and BP. AP and BP are
the required tangents. Thus length of the tangents are PA = PB =
7.4 cm.
Verification :
In the right angle triangle OAP , PA2 = OP2 −OA2 =
64 – 9 = 55
PA
= √55 = 7 4. cm (approximately) .
1. Prove that the
length of tangents drawn from an external point to a circle are equal.
Given:
A circle
with centre O; PA and PB are two tangents to the circle drawn from an external
point P.
To prove:
PA = PB
Construction:
Join OA,
OB, and OP.
It is
known that a tangent at any point of a circle is perpendicular to the radius
through the point of contact.
OA 
PA
and OB PB ...
(1)
In 
OPA
and OPB:
OAP
= OBP (Using
(1))
OA =
OB (Radii of the same circle)
OP =
OP (Common side)
Therefore, OPA 
OPB (RHS
congruency criterion)
PA = PB
(Corresponding
parts of congruent triangles are equal)
Thus, it
is proved that the lengths of the two tangents drawn from an external point to
a circle are equal.
2. prove that
in two concentric circles the chord of the larger circle which touches the
smaller circle is bisected at the point of contact
solution:
Let there is a circle having center O
Let AB is the tangent to the smaller circle and
chord to the larger circle.
Let P is the point of contact.
Now, draw a perpendicular OP to AB
Now, since AB is the tangent to the smaller circle,
So, ∠OPA = 90
Now, AB is the chord of the larger circle and OP is
perpendicular to AB.
Since the perpendicular drawn from the center of the
circle to the chord bisect it.
So, AP = PB
Hence, in two concentric circles, the chord of the
larger circle which touches the smaller circle is bisected at the point of
contact.
3. Two tangents TP and TQ are drawn to a circle
with center O from an external point T. Prove that angle PTQ = 2 OPQ.
We
know that, the lengths of tangents drawn from an external point to a
circle are equal.
∴ TP = TQ
In
ΔTPQ,
TP
= TQ
⇒ ∠TQP = ∠TPQ ...(1) (In a
triangle, equal sides have equal angles opposite to them)
∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)
∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))
⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)
We
know that, a tangent to a circle is perpendicular to the radius through
the point of contact.
OP
⊥ PT,
∴ ∠OPT = 90º
⇒ ∠OPQ + ∠TPQ = 90º
⇒ ∠OPQ = 90º – ∠TPQ
⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)
From
(1) and (2), we get
∠PTQ = 2∠OPQ