6.9 Circles Part 1
In the previous lessons we learnt about straight
lines, figures formed
by straight lines. Are there
figures other than these?
What about our universe?,
The earth goes around in the ecliptic path. Moon follows a circular path to go
around earth.
We also come across objects such as coin, wheels, cycle
tires, rings. They are all in circular shapes. In this lesson we shall study
their properties.
6.9.1
Definitions
Definitions:
Figure 
Definition 

‘Circle’ is a closed curve on a plane, with every point on the
curve at equal distance from a fixed point. The fixed point is called center
of the circle and is denoted by O. In the
figure, points P,A,Q,R and S are at same distance from O. 

‘Radius’ is the line segment joining the centre of the circle
with any point on the circle. It is denoted by r. There are many radii. In the figure OP,OQ,OA are radii. OP=OA=OQ. 

‘Chord’ is a line segment joining any two points on the circle.
In the figure, AQ and RS are two chords. 

‘Diameter’ is the line segment
passing through the center of the circle and having its end points on the
circle. It is denoted by d. It is the longest chord of the circle. In the figure, PQ is the diameter
and it passes through the center O. There are many diameters Note that d=PQ= PO+OQ =r+r =2r 

‘Circumference’ is the distance around the
circle (perimeter of the circle). In the figure, the distance measured from
point P to P through the points A, Q, S and R is the circumference. 

Circles
having same center but different radii are called ‘Concentric
circles’. C1, C2 and C3
are 3 circles with different radii OA, 

Different
circles having same radii are called ‘Congruent
circles’. C1 and C2 are 2 circles having same
radii OA(OA= 

‘Arc’
is a part of the circle. The curve RS is
an arc. 

‘Segment’ is a part of the region, bounded by the chord and
the arc. In the figure, RXSR
is a segment. A
chord divides the circle into two parts. Correspondingly we have two
segments: minor and major segments. 

ASBA is a ‘minor’ segment. (Region
bounded by the minor arc ASB and the chord
AB) 

ASBOA and ACBOA are ‘semi’ circles (regions bounded by equal arcs
ASB,ACB and the diameter AB). 

ASBA
is a ‘major’ segment. (Region
bounded by the larger(major) arc ASB and
the chord AB) 
6.9.2 Properties (Theorems):
6.9.2.1. In
a circle, the perpendicular from the center to the chord bisects the chord.
Data: In the adjoining figure, O is the center, AQ
is the chord and
To Prove: AB=BQ
Solution:
Steps 
Statement

Reason 


Consider the _{}OAB and _{}OQB 

1 
OA = OQ 
Radii
of the circle 

2 
_{}OBA =_{}OBQ=90^{0} 
It
is given that 

3 



4 
_{}OAB _{} _{}OQB 
SAS postulate 

5 
AB=BQ 
Corresponding sides are equal 
Conversely,
the line joining the mid point of a chord to the center is perpendicular to the
chord
Exercise: Prove yourself (Proof is similar to the above)
6.9.2.2. Equal chords of a circle
are equidistant from the center
Data: In the adjoining figure, O is the center, PQ
and RS are 2 equal chords. OX and OY are perpendiculars to PQ and RS
respectively.
To Prove: OX=OY
Construction: Join OP and OR
Solution:
Steps 
Statement

Reason 


Consider the _{}OPX and _{}ORY 

1 
2PX=PQ 
The
perpendicular OX bisects the chord PQ 

2 
2RY=RS 
The
perpendicular OY bisects the chord RS 

3 
PQ=RS 
Given
that chords are equal 

4 
_{}2PX=2RY i.e. PX=RY 
Substitute
vales from Step1 and Step2 in Step3 

5 
OP =OR 
Radii
of the circle 

6 
_{}PXO =_{}OYR=90^{0} 


4 
_{}PXO _{} _{}RYO 
SAS postulate 

5 
OX=OY 
Corresponding sides are equal 
6.9.2.3. Chords of a circle which are
equidistant from the center are equal.
This property is converse of 6.9.2.2.
Data:
In the adjoining figure, O is the center, PQ and RS are the chords. OX
and OY are perpendicular to PQ and RS respectively and OX=OY To
Prove: PQ=RS Construction:
Join OP and OR Exercise: Follow the steps described above (6.9.2.2) to prove that _{}PXO _{} _{}RYO and then show that PX=RY 

6.9.2 Problem 1: In the adjoining figure AB and CD
are equal chords of a circle whose center is O, when produced these chords
meet at E. Prove that EB=ED
Construction: Draw Perpendiculars to AB and CD to
meet AB at P and CD at Q, Join OE
Steps 
Statement

Reason 

1 
AP =1/2AB 
Perpendicular bisects the chord 

2 
CQ= 1/2CD 
Perpendicular bisects the chord 

3 
AP=CQ 
AB=CD(given) 

4 
_{}OPB =_{}OQD = 90^{0} 
Construction 

5 
OP =OQ 
Equal chords are equidistant. 

6 
OE is common 
Construction 

7 
_{}OPE _{} _{}OPB 
RHS postulate on congruence 

8 
PE=QE 
Corresponding sides are equal 

9 
AP+PE =CQ+QE 
Step 3 and 8 
6.9.2 Theorem (Inscribed Angle Theorem): In any circle, the angle subtended by an arc
at the center of the circle is double the
angle subtended by the same arc at any point on the remaining part of the
circle.
Data: In the adjoining figure, O is the center of
the circle. _{}AOB is the angle subtended by the arc AXB, at the centre of
the circle.
_{}APB is the angle subtended by the same arc at any point (P) on
the remaining part of the circle.
To prove: _{}AOB = 2_{}APB
Construction: Extend
Proof:
Steps 
Statement

Reason 

1 
OA = OP 
Radii of the circle 

2 
_{}OPA = _{}OAP 
_{}OAP is an isosceles triangle 

3 
_{}AOD =_{}OAP+_{}OPA 
Exterior angle in a triangle (_{}AOP) is equal to the sum of two interior angles. 

4 
= 2_{}OPA 
Substitute result of Step 2 in
Step 3. 

5 

Radii of circle 

6 
_{}OBP = _{}OPB 
_{}OBP is an isosceles triangle 

7 
_{}BOD =_{}OBP+_{}OPB 
Exterior angle in a triangle (_{}BOP) is equal to the sum of two interior angles. 

8 
= 2_{}OPB 
Substitute the result from Step 6
in Step 7 

9 
_{}AOD+ _{}BOD=2_{}OPA+2_{}OPB =2(_{}OPA+_{}OPB)= 2_{}APB 
From steps 3,4,7 and 8 

10 
_{}_{}AOB =2_{}APB 
_{}AOB=_{}AOD+ _{}BOD 
Corollary : Angle in a semi circle
is a right angle
In
the adjacent figure AB is radius and _{}ACB is the angle at semi circle, we
are required to prove that_{} ACB = 90^{0} ^{ } Hint: From
the above theorem, 2_{} ACB = _{}AOB Since AOB is a straight line, it follows that _{} AOB = 180^{0} Hence
_{} ACB = 90^{0} 

Theorem: Prove that inscribed
angles in the same segment of a circle are equal
Given: A and B are points on the circle. ACB and
ADB are the inscribed angles and AOB is the central angle
To Show: _{}ACB= _{}ADB
Steps 
Statement

Reason 

1 
_{}AOB= 2_{}ADB 
Central
angle is double the inscribed angle of the same segment (arc) AB 

2 
_{}AOB= 2_{}ACB 
Central
angle is double the inscribed angle of the same segment (arc) AB 

3 
_{} 2_{}ACB= 2_{}ADB 
Equating
results from steps 1 and 2 

4 
i.e.
_{}ACB= _{}ADB 

Converse of the above theorem is the below
mentioned theorem:
Theorem: (Proof not provided)
If a line segment joining two points subtends equal
angles at two other points lying on the same side of the line containing the
segment,
then the four points lie on
the same circle.(In the above figure ABCD will be concyclic, if _{}ACB= _{}ADB)
The proof is by logical reasoning. (Making an assumption that the theorem is not
true and then proving that assumption made is wrong)
6.9.2 Problem 2: Prove that the _{}APC and _{}DPB are equiangular in the adjoining figure
Also prove that the product of their segments is
equal
Given: AC and BD are 2 chords of the same circle
To Show: _{}APC and _{}DPB are equiangular and PC*PD =BP*PA
Steps 
Statement

Reason 

1 
_{}ACD = _{}ABD 
Inscribed
angles on the circle formed by the same segment (arc) AD 

2 
_{}CAB = _{}CDB 
Inscribed
angles on the circle formed by the same segment (arc) BC 

3 
_{}CPA = _{}BPD 
Vertically
opposite angles 

4 
_{}APC _{}DPB 
AAA postulate on similarity 

5 
AC/BD
= PD/PA =PB/PC 
corresponding
sides are proportional 

6 
PC*PD
=PA*PB 

Note This
proves the following theorem
Theorem : If two chords of a circle intersect internally or
externally then the product of the lengths of their segments are equal.
On the lines of 6.9.2 Problem 2 we can
prove the above theorem when P is outside the circle also.
6.9.2 Problem 3: In the adjoining figure, AB and BC
are diameters of two circles intersecting at B and D. Show that A, D and C are collinear.
Given: AB and BC are diameters.
To Show: _{}ADC = 180^{0}
Steps 
Statement

Reason 

1 
_{}APB = 180^{0} 
Angle
on the straight line 

2 
_{}APB = 2_{}ADB 
Central
angle is double the inscribed angle of the same segment (arc) AB 

2 
_{} 2_{}ADB = 180^{0} 
From
Step1 and Step 2 

3 
i.e.
_{}ADB = 90^{0} 


4 
_{}BQC = 180^{0} 
Angle
on the straight line 

5 
_{}BQC = 2_{}BDC 
Central
angle is double the inscribed angle of the same segment (arc) BC 

6 
_{} 2_{}BDC = 180^{0} 
From
Step4 and Step 5 

7 
i.e.
_{}BDC = 90^{0} 


8 
_{}ADB+_{}BDC = 180^{0} 
From
Step 3 and Step 7 
6.9.2 Problem 4: In the adjoining figure, two
circles intersect at two points B and C. Through B, two line segments ABD and
PBQ are
drawn to intersect the
circles at A,D and P,Q respectively. Prove that _{}ACP = _{}QCD
Steps 
Statement

Reason 

1 
_{}ACP = _{}ABP ^{} 
Angle
on the circle on the same chord AP 

2 
_{}ABP = _{}DBQ 
Vertically
opposite angles. 

3 
_{}DBQ = _{}DCQ 
Angle
on the circle on the same chord DQ 

4 
_{}ACP = _{}QCD 
From
Step 1,2,3 
6.9.2 Problem 5: In the adjoining figure, the
bisector of _{}B of an isosceles triangle ABC with AB=AC meets the
circumcircle of _{}ABC at P.
If AP and BC produced meet at Q, prove that CQ=CA
Construction: Join CP.
Steps 
Statement

Reason 

1 
_{}ABC = 2_{}CBP 
BP is bisector of _{}ABC 

2 
_{}CBP=_{}CAP 
The
chord PC subtends same angle 

3 
_{}ABC = 2_{}CAP 
Step
1 and 2 

4 
_{}BCA = _{}CAQ+_{}CQA 
Exterior
angle in a _{} is sum of interior
angles 

5 
_{}CQA = _{}BCA_{}CAP 
Simplification 

6 
=
_{}ABC_{}CAP 
_{}BCA =_{}ABC as AB=AC 

7 
=2_{}CAP_{}CAP = _{}CAP 
Step
3 

8 
CQ=CA

Step5,6(Angles
on the base AQ are equal) 
Construction of Circle:
1. Can we construct a unique circle given just a point? No, because we can construct
several circles passing through a point 2. Can we construct a unique circle given two noncollinear points? No, because we can construct
several circles passing through two points Note in the adjacent figure
through P and Q we can construct several circles 3. Can we construct a unique circle given three noncollinear points? yes. We could do that
and steps are as follows: 

Method:
Steps 
construction 

1 
Take 3 points A, B, C 

2 
Join AB, BC 

3 
Construct perpendicular
bisectors to AB,BC(Refer 6.4.3) 

4 
Let these bisectors meet at S 

5 
With SA as radius draw circle 
Note that this circle touches B and C
as well; In fact this circle is circumcircle of _{} ABC and S is its Circumcenter
Note: Refer Section 6.4.3 to
know more about circumcircle.
Since SA=SB=SC it is
proved that S is the center of circle, passing through points A,B,C.
This proves that there
is only one circle which passes through three points which are not collinear.
4. Construction of a
circle (need not be unique) given two
points.
Method:
Let A and B are the
given two points.
Let C be a point on the
plane such that A, B and C are noncollinear.
Draw perpendicular
bisectors of AB and BC as above and let they meet at S.
The circle with S as
center and SA as radius passes through B.
6.9.3
Cyclic quadrilateral
Definition: A quadrilateral whose
vertices lie on a circle is called a ‘cyclic
quadrilateral’. It is an inscribed (inside a circle) quadrilateral.
6.9.3 Theorem: Opposite angles of a cyclic quadrilateral are
supplementary (i.e. their sum is 180^{0}).
Data: ABCD is a cyclic quadrilateral and O is the
center of circle.
To prove: _{}BAD + _{}BCD = 180^{0} and _{}ABC +_{}ADC = 180^{0}.
Construction: Join
Proof:
Steps 
Statement

Reason 

1 
_{}BAD = 1/2 of _{}BOD 
Inscribed angle is half the
angle at center 

2 
_{}BCD = 1/2 of reflex _{}BOD 
Inscribed angle is half the
angle at center 

3 
_{}_{}BAD +_{}BCD = 1/2 _{}BOD + 1/2 of reflex_{}BOD = 1/2(_{}BOD+ reflex_{}BOD) = 1/2(360^{0}) = 180^{0} 
(_{}BOD+ reflex_{}BOD) is the complete angle at the center and is equal to = 360^{0} ^{ } 

4 
Similarly _{}ABC +_{}ADC = 180^{0} 

This proves that
opposite angles of a cyclic quadrilateral are supplementary (i.e. their sum is
180^{0}).
Note :
When we join BD, then
we get two segments:BAD as a major segment and BCD as
a minor segment on the same chord BD.
They are also referred
as ‘alternate segments’
The above theorem can
be restated as
Theorem: Angles in the alternate segments of
a circle are supplementary.
Converse of above Theorem :( Proof not provided): If the opposite angles of a
quadrilateral are supplementary, then it is cyclic.
The proof is by logical reasoning. (Making an assumption that the theorem is not
true and then proving that assumption made is wrong)
6.9.3 Problem 1: Prove that the exterior angle of a cyclic
quadrilateral is equal to its interior opposite angle.
Given: ABCD is a cyclic
quadrilateral. _{}DCE is the exterior angle
To prove: _{}BAD =_{}DCE
Steps 
Statement

Reason 

1 
_{}BAD+ _{}BCD = 180^{0} 
Opposite angles of
a cyclic quadrilateral is supplementary 

2 
_{}BCD + _{}DCE = 180^{0} 
Linear pair or adjacent angles
on a straight line 

3 
_{}BAD+ _{}BCD = _{}BCD + _{}DCE _{}_{}BAD =_{}DCE 
Subtract _{}BCD from both sides 
This proves that the exterior angle of a
cyclic quadrilateral is equal to its interior opposite angle.
6.9.3 Problem 2: Prove that when a parallelogram is inscribed in a
circle it becomes a rectangle
Given: ABCD is a
parallelogram and ABCD is a cyclic quadrilateral
To prove: _{}ABC = _{}BCD = _{}ADC = _{}DAB = 90^{0}
Steps 
Statement 
Reason 

1 
_{}BAD+ _{}BCD = 180^{0} 
Opposite
angles of a cyclic quadrilateral is supplementary 

2 
_{}BAD = _{}BCD 
Opposite
angles in a parallelogram are equal.(Refer 6.7) 

3 
_{}BAD =_{}BCD =90^{0} 
From
step 1,2 
This proves that when parallelogram ABCD is inscribed in a circle it becomes a
rectangle
6.9.4
Construction of a cyclic Quadrilateral
Steps to be followed (general):
Note: We need to draw a
circle which passes through all the 4 vertices of a quadrilateral. We have
learnt that the circumcircle of a triangle
passes through all the 3
vertices of a triangle. So our problem will be solved if we can construct a circumcircle
and then locate a point on
that circle which is the
fourth vertex of the quadrilateral.
Step 1: Construct a
Triangle with the given data
Step 2: Bisect any two
sides of this Triangle (To find the Circumcenter)
Step 3: Join these
bisectors to meet at origin O
Step 4: With O as
origin, draw a circle passing through 3 points of the triangle drawn in Step1
Step 5: Cut an arc of
given length on the circle, to locate the 4^{th} point.
Note: To construct a
triangle we need three values (elements). They could be any of the following:
1. Length of three
sides
2. Length of one side
and two angles on this line
3. Length of two sides
and the included angle
6.9.4 Problem 1:
Construct a cyclic quadrilateral
KLMN with KL = 4cm, LM = 4.8cm, KM = 6.8cm and KN = 4.3cm.
Steps:
1. Construct the triangle KLM as
follows: (i)
Draw the line KL=4cm. (ii) From K, draw an arc of radius of 6.8cm. From
L, draw another
arc of radius 4.8cm. Let they meet at M. (iii) Join KM and LM to form the triangle
KLM. 2. Bisect the lines KL and LM
and extend the bisector lines to meet at point O 3. With O as origin, draw a
circle passing through the points K, L and M 4. From K, draw an arc of radius
4.3cm to cut the circle at point N 5. Draw the quadrilateral KLMN 

6.9.4 Problem 2:
Construct a cyclic
quadrilateral XYZT with XY= 2.5cm, YZ=5.5cm, ZT=3cm and _{}XTZ = 60^{0}.
Note: Since XYZT is a
cyclic quadrilateral, the opposite angles _{}XTZ and _{}XYZ are supplementary and hence _{}XYZ= 120^{0} .
Steps:
1. Construct the triangle XYZ as
follows: (i)
Draw the line XY=2.5cm. (ii)
From Y, draw a line at an angle 120^{0} with XY. Draw an arc
of radius 5.5cm from Y to cut this line at Z (iii) Join XZ to form the triangle XYZ. 2. Bisect the lines XY and YZ
and extend the bisector lines to meet at point O 3. With O as origin draw a
circle passing through the points X, Y and Z 4. From Z, draw an arc of radius
3cm to cut the circle at point T 5. Draw the quadrilateral XYZT. 

6.9.2.4. Equal chords of a circle subtend equal angles at
the center
Proof:
Steps 
Statement

Reason 


Consider the _{}AOB and _{}COD 

1 
OA = OD 
Radii 

2 

Radii 

3 
AB = CD 
Given 

4 
_{}AOB _{}_{}COD 
SSS postulate 

5 
_{}AOB =_{}COD 
Corresponding
angles are equal 
6.9.2.5. If the angles
subtended by chords of a circle at the center are equal then these chords are
equal
Note: This is converse of 6.9.2.4.
Proof:
Steps for the proof are similar to the given in
section 6.9.2.4, except that in the third step, use the given data that _{}AOB =_{}COD
and then by SAS property, prove that the
triangles are congruent and hence show AB=CD.
Note: There is one and only one
circle passing through 3 noncollinear points.
6.9 Summary of learning
No 
Points to remember 
1 
In a circle, the perpendicular from the center to
the chord bisects the chord. 
2 
Equal chords of a circle are equidistant from the
center 
3 
Chords of a circle which are equidistant from the
center are equal. 
4 
In any circle, the angle subtended by an arc at
the center is double the angle subtended by the same arc at any point on the
remaining part of the circle 
5 
Opposite angles of a cyclic
quadrilateral are supplementary (i.e. their sum is 180^{0}). 
Additional points:
6.9.5 Area of circle
The length around a circle or the perimeter of a
circle is called its ‘circumference’. If ‘r’ is
the radius of any circle, then the formula for the circumference is c = 2_{}r , where _{} is a constant. We
use its approximate value of 22/7 in all our calculations. The
area of a circle = _{}r^{2}. 

Observation:
When we discuss _{}, the names which
come to our mind are the mathematicians Aryabhata and Bhaskaracharya
1)
Aryabhatta of 5^{th} century AD was the first one
to give approximate value of _{} to 4 correct decimal
places (3.1416).
His formula is:
The approximate
circumference of a circle of diameter 20000
units is got by adding 62000 to the result of 8 times the sum of 100 and 4.
Circumference = 62000+ 8(100+4) =
62832; Diameter = 200
_{} _{}= circumference ÷ diameter = 62832 ÷ 20000= 3.1416
2)
Bhaskarachary’s formula ( Lilavati, Shloka 202)
The circumference (approximate value) of a circle
is got by multiplying its diameter by 3927 and then dividing the product
by1250.
For simpler calculations, the circumference of a
circle is got by multiplying its diameter by 22 and then dividing the product
by 7.
_{} = circumference ÷
diameter = 3927 ÷ 1250 = 3.1416
For simpler calculations, _{} = 22/7
6.9.5 Problem 1: Area of two circles are
in the ratio of 25:36. Find the ratio of their circumferences.
Solution:
If r and R are the radii
of the two circles then their areas are _{}r^{2} and _{}R^{2}.
It is given that _{}r^{2}:_{}R^{2}= 25:36
_{} r^{2}:R^{2}=
25:36 = 5^{2}:6^{2}
_{} r:R
= 5:6
_{} 2_{}r:2_{}R = 2_{}*5:2_{}*6 = 5:6
6.9.5 Problem 2: A well of diameter 150cm has a 30cm wide parapet
running around it. Find the area of the parapet.
Solution:
Since
the diameter of the well is 150cm, it’s radius =
75cm. Well
can be imagined as a circle C_{1} having of radius OA(r) = 75cm as
shown in the adjoining figure. C_{2}
is another circle around the well with a radius of OB(R) = (75+30)cm = 105cm Area
of the circle C_{2 }= _{}R^{2 }= 22/7*105*105 = 34650 sq.cm. Area
of the circle C_{1 }= _{}r^{2 }= 22/7*75*75 = 17678.6 sq.cm. _{} Area of the parapet
= _{}R^{2}_{}r^{2}= 3465017678.6 = 16971.4
sq.cm. = 1.7 sq.m. 

Brahmagupta’s theorem:
A cyclic quadrilateral in which the diagonals are
perpendicular to each other, the perpendicular through the point of
intersection of the
diagonals to one of the sides,
bisects the opposite side. Though we do not know his proof, our proof based on
what has been learnt
so far is as follows.
Given: ABCD is a cyclic
quadrilateral. AC_{}BD, MG _{}CD, GM produced meets AB at H.
To prove: AH=BH
Steps 
Statement 
Reason 


Consider the _{}MDG and _{}MAH 

1 
_{}MDG + _{}GMD = 90^{0} 
_{}MGD = 90^{0 }(given) and sum of angles in a
triangle is 180^{0} 

2 
_{}AMH + _{}BMH = 90^{0} 
Given
that diagonals are perpendicular to each other 

3 
_{}BMH =_{}GMD 
Vertically
opposite angles 

4 
_{}MDG = _{}AMH 
Steps
1,2,3 

5 
_{}BDC = _{}MDG = _{}BAM 
Inscribed
angles on same chord BC 

6 
_{}BAM = _{}AMH 
Steps
4,5 

7 
AH = HM 
Step 6
(AMH is an isosceles triangle) 


Consider the _{}MCG and _{}MBH 

8 
_{}MCG + _{}GMC = 90^{0} 
_{}MGC = 90^{0}(given) and sum of angles in a triangle
is 180^{0} 

9 
_{}AMH + _{}BMH = 90^{0} 
Given
that diagonals are perpendicular to each other 

10 
_{}AMH =_{}GMC 
Vertically
opposite angles 

11 
_{}MCG = _{}BMH 
Steps
8,9,10 

12 
_{}ABD = _{}MCG = _{}ABM 
Inscribed
angles on same chord AD 

13 
_{}ABM = _{}BMH 
Steps
11,12 

14 
BH = HM 
Step 6
(AMH is an isosceles triangle) 

15 
AH = BH 
Steps
7 and 14 