Detailed Notes: Circles | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


CIRCLE 
Imagine you have a single point in a wide, empty space. This point is the center of a 
circle. A circle is formed by drawing a line that keeps the same distance all around from 
this center point. Mathematicians love circles because they're simple yet full of surprises. 
In math, a circle is all the points in a plane that are the same distance from a fixed point, 
the center. This fixed distance is called the radius. Circles are cool because they can be 
described in many ways using math equations. Let's explore these equations and look at 
some interesting problems that show off circles' special properties. 
Equation of a circle in various forms: 
(a) The circle with center as origin & radius ' ?? ' has the equation; ?? 2
+ ?? 2
= ?? 2
. 
 
(b) The circle with centre ( h, ?? ) & radius ' ?? ' has the equation; ( ?? - h)
2
+ ( ?? - ?? )
2
= ?? 2
 
(c) The general equation of a circle is ?? 2
+ ?? 2
+ 2???? + 2???? + ?? = 0 
with centre as ( -?? , -?? ) & radius = v?? 2
+ ?? 2
- ?? . 
This can be obtained from the equation ( ?? - h)
2
+ ( ?? - ?? )
2
= ?? 2
 
? ?? 2
+ ?? 2
- 2h?? - 2???? + h
2
+ ?? 2
- ?? 2
= 0 
Take  -h = ?? , -?? = ?? and h
2
+ ?? 2
- ?? 2
= ?? 
Condition to define circle :- 
?? 2
+ ?? 2
- ?? > 0 ? real circle. 
?? 2
+ ?? 2
- ?? = 0 ? point circle. 
?? 2
+ ?? 2
- ?? < 0 ? imaginary circle, with real centre, that is ( -?? , -?? ) 
Page 2


CIRCLE 
Imagine you have a single point in a wide, empty space. This point is the center of a 
circle. A circle is formed by drawing a line that keeps the same distance all around from 
this center point. Mathematicians love circles because they're simple yet full of surprises. 
In math, a circle is all the points in a plane that are the same distance from a fixed point, 
the center. This fixed distance is called the radius. Circles are cool because they can be 
described in many ways using math equations. Let's explore these equations and look at 
some interesting problems that show off circles' special properties. 
Equation of a circle in various forms: 
(a) The circle with center as origin & radius ' ?? ' has the equation; ?? 2
+ ?? 2
= ?? 2
. 
 
(b) The circle with centre ( h, ?? ) & radius ' ?? ' has the equation; ( ?? - h)
2
+ ( ?? - ?? )
2
= ?? 2
 
(c) The general equation of a circle is ?? 2
+ ?? 2
+ 2???? + 2???? + ?? = 0 
with centre as ( -?? , -?? ) & radius = v?? 2
+ ?? 2
- ?? . 
This can be obtained from the equation ( ?? - h)
2
+ ( ?? - ?? )
2
= ?? 2
 
? ?? 2
+ ?? 2
- 2h?? - 2???? + h
2
+ ?? 2
- ?? 2
= 0 
Take  -h = ?? , -?? = ?? and h
2
+ ?? 2
- ?? 2
= ?? 
Condition to define circle :- 
?? 2
+ ?? 2
- ?? > 0 ? real circle. 
?? 2
+ ?? 2
- ?? = 0 ? point circle. 
?? 2
+ ?? 2
- ?? < 0 ? imaginary circle, with real centre, that is ( -?? , -?? ) 
Note : That every second degree equation in ?? &?? , in which coefficient of ?? 2
 is equal to 
coefficient of ?? 2
 & the coefficient of ???? is zero, always represents a circle. 
(d) The equation of circle with ( ?? 1
, ?? 1
) &( ?? 2
, ?? 2
) as extremeties of its diameter is: 
( ?? - ?? 1
) ( ?? - ?? 2
)+ ( ?? - ?? 1
) ( ?? - ?? 2
) = 0. 
 
This is obtained by the fact that angle in a semicircle is a right angle. 
?  (Slope of PA) (Slope of PB) = -1 
?
?? - ?? 1
?? - ?? 1
·
?? - ?? 2
?? - ?? 2
= -1 ? ( ?? - ?? 1
) ( ?? - ?? 2
)+ ( ?? - ?? 1
) ( ?? - ?? 2
) = 0 
Note that this will be the circle of least radius passing through ( ?? 1
, ?? 1
) &( ?? 2
, ?? 2
) . 
Problem 1 Find the equation of the circle whose centre is ( 0,3) and radius is 3 . 
Solution. The equation of the circle is ( ?? - 0)
2
+ ( ?? - 3)
2
= 3
2
 
? ?? 2
+ ?? 2
- 6?? = 0 
Problem 2 Find the equation of the circle which passes through ( 1, -1) and two of its 
dimeter are ?? + 2?? - 5 = 0 and ?? - ?? + 1 = 0 
Solution. Let ?? be the point of intersection of the lines 
?? + 2?? - 5 = 0 
and ?? - ?? + 1 = 0 
Solving (i) and (ii), we get ?? = 1, ?? = 2. So, coordinates of centre are ( 1,2) . since circle 
passes through ( 1, -1) so 
radius = v( 1 - 1)
2
+ ( 2 + 1)
2
 ?  radius = 3 
Hence the equation of the required circle is ( ?? - 1)
2
+ ( ?? + 2)
2
= 9 
Problem 3 If the equation ?? ?? 2
+ ( ?? - 3) ???? + 3?? 2
+ 6???? + 2???? - 3 = 0 represents the 
equation of a circle then find ?? , ?? 
Page 3


CIRCLE 
Imagine you have a single point in a wide, empty space. This point is the center of a 
circle. A circle is formed by drawing a line that keeps the same distance all around from 
this center point. Mathematicians love circles because they're simple yet full of surprises. 
In math, a circle is all the points in a plane that are the same distance from a fixed point, 
the center. This fixed distance is called the radius. Circles are cool because they can be 
described in many ways using math equations. Let's explore these equations and look at 
some interesting problems that show off circles' special properties. 
Equation of a circle in various forms: 
(a) The circle with center as origin & radius ' ?? ' has the equation; ?? 2
+ ?? 2
= ?? 2
. 
 
(b) The circle with centre ( h, ?? ) & radius ' ?? ' has the equation; ( ?? - h)
2
+ ( ?? - ?? )
2
= ?? 2
 
(c) The general equation of a circle is ?? 2
+ ?? 2
+ 2???? + 2???? + ?? = 0 
with centre as ( -?? , -?? ) & radius = v?? 2
+ ?? 2
- ?? . 
This can be obtained from the equation ( ?? - h)
2
+ ( ?? - ?? )
2
= ?? 2
 
? ?? 2
+ ?? 2
- 2h?? - 2???? + h
2
+ ?? 2
- ?? 2
= 0 
Take  -h = ?? , -?? = ?? and h
2
+ ?? 2
- ?? 2
= ?? 
Condition to define circle :- 
?? 2
+ ?? 2
- ?? > 0 ? real circle. 
?? 2
+ ?? 2
- ?? = 0 ? point circle. 
?? 2
+ ?? 2
- ?? < 0 ? imaginary circle, with real centre, that is ( -?? , -?? ) 
Note : That every second degree equation in ?? &?? , in which coefficient of ?? 2
 is equal to 
coefficient of ?? 2
 & the coefficient of ???? is zero, always represents a circle. 
(d) The equation of circle with ( ?? 1
, ?? 1
) &( ?? 2
, ?? 2
) as extremeties of its diameter is: 
( ?? - ?? 1
) ( ?? - ?? 2
)+ ( ?? - ?? 1
) ( ?? - ?? 2
) = 0. 
 
This is obtained by the fact that angle in a semicircle is a right angle. 
?  (Slope of PA) (Slope of PB) = -1 
?
?? - ?? 1
?? - ?? 1
·
?? - ?? 2
?? - ?? 2
= -1 ? ( ?? - ?? 1
) ( ?? - ?? 2
)+ ( ?? - ?? 1
) ( ?? - ?? 2
) = 0 
Note that this will be the circle of least radius passing through ( ?? 1
, ?? 1
) &( ?? 2
, ?? 2
) . 
Problem 1 Find the equation of the circle whose centre is ( 0,3) and radius is 3 . 
Solution. The equation of the circle is ( ?? - 0)
2
+ ( ?? - 3)
2
= 3
2
 
? ?? 2
+ ?? 2
- 6?? = 0 
Problem 2 Find the equation of the circle which passes through ( 1, -1) and two of its 
dimeter are ?? + 2?? - 5 = 0 and ?? - ?? + 1 = 0 
Solution. Let ?? be the point of intersection of the lines 
?? + 2?? - 5 = 0 
and ?? - ?? + 1 = 0 
Solving (i) and (ii), we get ?? = 1, ?? = 2. So, coordinates of centre are ( 1,2) . since circle 
passes through ( 1, -1) so 
radius = v( 1 - 1)
2
+ ( 2 + 1)
2
 ?  radius = 3 
Hence the equation of the required circle is ( ?? - 1)
2
+ ( ?? + 2)
2
= 9 
Problem 3 If the equation ?? ?? 2
+ ( ?? - 3) ???? + 3?? 2
+ 6???? + 2???? - 3 = 0 represents the 
equation of a circle then find ?? , ?? 
Solution.  ?? ?? 2
+ ( ?? - 3) ???? + 3?? 2
+ 6???? + 2???? - 3 = 0 
above equation will represent a circle if 
coefficient of ?? 2
= coefficient of ?? 2
 
?? = 3 
coefficient of ???? = 0 
?? = 3 
Problem 4 Find the equation of a circle whose diametric end points are ( ?? 1
, ?? 1
) and 
( ?? 2
, ?? 2
) where ?? 1
, ?? 2
 are the roots of ?? 2
- ???? + ?? = 0 and ?? 1
, ?? 2
 are the roots of ?? 2
- ???? +
?? = 0. 
Solution. We know that the equation of the circle described on the line segment joining 
( ?? 1
, ?? 1
) and ( ?? 2
, ?? 2
) as a diameter is ( ?? - ?? 1
) ( ?? - ?? 2
)+ ( ?? - ?? 1
) ( ?? - ?? 2
) = 0. 
?? 2
+ ?? 2
- ( ?? 1
+ ?? 2
) ?? - ( ?? 1
+ ?? 2
) ?? + ?? 1
?? 2
+ ?? 1
?? 2
= 0 
Here, 
 ?? 1
+ ?? 2
= ?? , ?? 1
?? 2
= ??  ?? 1
+ ?? 2
= ?? , ?? 1
?? 2
= ??  
So, the equation of the required circle is 
?? 2
+ ?? 2
- ???? - ???? + ?? + ?? = 0 
Intercepts made by a circle on the axes: 
Intercepts are the points where the axes meet the circle. 
The intercepts made by the circle ?? 2
+ ?? 2
+ 2???? + 2???? + ?? = 0 on the co-ordinate axes 
are 2v?? 2
- ?? (on ?? -axis) &2v?? 2
- ?? (on ?? -axis) respectively. 
If, 
?? 2
> ?? ?  ???????????? ???????? ?? h?? ?? ???????? ???? ?????? ???????????????? ???????????? .  ?? 2
= ?? 
?  ???????????? ???????? h???? ?? h?? ?? - ???????? .  ?? 2
< ?? 
?  ???????????? ???????? ???????????????????? ?????????? ???? ?????????? ?? h?? ?? - ???????? .   
Page 4


CIRCLE 
Imagine you have a single point in a wide, empty space. This point is the center of a 
circle. A circle is formed by drawing a line that keeps the same distance all around from 
this center point. Mathematicians love circles because they're simple yet full of surprises. 
In math, a circle is all the points in a plane that are the same distance from a fixed point, 
the center. This fixed distance is called the radius. Circles are cool because they can be 
described in many ways using math equations. Let's explore these equations and look at 
some interesting problems that show off circles' special properties. 
Equation of a circle in various forms: 
(a) The circle with center as origin & radius ' ?? ' has the equation; ?? 2
+ ?? 2
= ?? 2
. 
 
(b) The circle with centre ( h, ?? ) & radius ' ?? ' has the equation; ( ?? - h)
2
+ ( ?? - ?? )
2
= ?? 2
 
(c) The general equation of a circle is ?? 2
+ ?? 2
+ 2???? + 2???? + ?? = 0 
with centre as ( -?? , -?? ) & radius = v?? 2
+ ?? 2
- ?? . 
This can be obtained from the equation ( ?? - h)
2
+ ( ?? - ?? )
2
= ?? 2
 
? ?? 2
+ ?? 2
- 2h?? - 2???? + h
2
+ ?? 2
- ?? 2
= 0 
Take  -h = ?? , -?? = ?? and h
2
+ ?? 2
- ?? 2
= ?? 
Condition to define circle :- 
?? 2
+ ?? 2
- ?? > 0 ? real circle. 
?? 2
+ ?? 2
- ?? = 0 ? point circle. 
?? 2
+ ?? 2
- ?? < 0 ? imaginary circle, with real centre, that is ( -?? , -?? ) 
Note : That every second degree equation in ?? &?? , in which coefficient of ?? 2
 is equal to 
coefficient of ?? 2
 & the coefficient of ???? is zero, always represents a circle. 
(d) The equation of circle with ( ?? 1
, ?? 1
) &( ?? 2
, ?? 2
) as extremeties of its diameter is: 
( ?? - ?? 1
) ( ?? - ?? 2
)+ ( ?? - ?? 1
) ( ?? - ?? 2
) = 0. 
 
This is obtained by the fact that angle in a semicircle is a right angle. 
?  (Slope of PA) (Slope of PB) = -1 
?
?? - ?? 1
?? - ?? 1
·
?? - ?? 2
?? - ?? 2
= -1 ? ( ?? - ?? 1
) ( ?? - ?? 2
)+ ( ?? - ?? 1
) ( ?? - ?? 2
) = 0 
Note that this will be the circle of least radius passing through ( ?? 1
, ?? 1
) &( ?? 2
, ?? 2
) . 
Problem 1 Find the equation of the circle whose centre is ( 0,3) and radius is 3 . 
Solution. The equation of the circle is ( ?? - 0)
2
+ ( ?? - 3)
2
= 3
2
 
? ?? 2
+ ?? 2
- 6?? = 0 
Problem 2 Find the equation of the circle which passes through ( 1, -1) and two of its 
dimeter are ?? + 2?? - 5 = 0 and ?? - ?? + 1 = 0 
Solution. Let ?? be the point of intersection of the lines 
?? + 2?? - 5 = 0 
and ?? - ?? + 1 = 0 
Solving (i) and (ii), we get ?? = 1, ?? = 2. So, coordinates of centre are ( 1,2) . since circle 
passes through ( 1, -1) so 
radius = v( 1 - 1)
2
+ ( 2 + 1)
2
 ?  radius = 3 
Hence the equation of the required circle is ( ?? - 1)
2
+ ( ?? + 2)
2
= 9 
Problem 3 If the equation ?? ?? 2
+ ( ?? - 3) ???? + 3?? 2
+ 6???? + 2???? - 3 = 0 represents the 
equation of a circle then find ?? , ?? 
Solution.  ?? ?? 2
+ ( ?? - 3) ???? + 3?? 2
+ 6???? + 2???? - 3 = 0 
above equation will represent a circle if 
coefficient of ?? 2
= coefficient of ?? 2
 
?? = 3 
coefficient of ???? = 0 
?? = 3 
Problem 4 Find the equation of a circle whose diametric end points are ( ?? 1
, ?? 1
) and 
( ?? 2
, ?? 2
) where ?? 1
, ?? 2
 are the roots of ?? 2
- ???? + ?? = 0 and ?? 1
, ?? 2
 are the roots of ?? 2
- ???? +
?? = 0. 
Solution. We know that the equation of the circle described on the line segment joining 
( ?? 1
, ?? 1
) and ( ?? 2
, ?? 2
) as a diameter is ( ?? - ?? 1
) ( ?? - ?? 2
)+ ( ?? - ?? 1
) ( ?? - ?? 2
) = 0. 
?? 2
+ ?? 2
- ( ?? 1
+ ?? 2
) ?? - ( ?? 1
+ ?? 2
) ?? + ?? 1
?? 2
+ ?? 1
?? 2
= 0 
Here, 
 ?? 1
+ ?? 2
= ?? , ?? 1
?? 2
= ??  ?? 1
+ ?? 2
= ?? , ?? 1
?? 2
= ??  
So, the equation of the required circle is 
?? 2
+ ?? 2
- ???? - ???? + ?? + ?? = 0 
Intercepts made by a circle on the axes: 
Intercepts are the points where the axes meet the circle. 
The intercepts made by the circle ?? 2
+ ?? 2
+ 2???? + 2???? + ?? = 0 on the co-ordinate axes 
are 2v?? 2
- ?? (on ?? -axis) &2v?? 2
- ?? (on ?? -axis) respectively. 
If, 
?? 2
> ?? ?  ???????????? ???????? ?? h?? ?? ???????? ???? ?????? ???????????????? ???????????? .  ?? 2
= ?? 
?  ???????????? ???????? h???? ?? h?? ?? - ???????? .  ?? 2
< ?? 
?  ???????????? ???????? ???????????????????? ?????????? ???? ?????????? ?? h?? ?? - ???????? .   
 
???? = 2???? = 2
v
?? 2
- ?? ?? 2
= 2v?? 2
- ?? 2
= 2v?? 2
+ ?? 2
- ?? - ?? 2
= 2v?? 2
- ?? 
Problem 5 Find the locus of the centre of the circle whose ?? and ?? intercepts are ?? and ?? 
respectively. 
Solution. Equation of circle is ?? 2
+ ?? 2
+ 2???? + 2???? + ?? = 0 ?? intercept = ?? 
2v?? 2
- ?? = ?? ?? 2
- ?? =
?? 2
4
 ?? ?????????????????? = ??  2v?? 2
- ?? = ?? ?? 2
- ?? =
?? 2
4
 #( ???? )  
subtracting equation (i) and (ii) 
?? 2
- ?? 2
=
?? 2
- ?? 2
4
 
hence locus of centre is ?? 2
- ?? 2
=
?? 2
-?? 2
4
 
Parametric equations of a circle: 
The parametric equations of ( ?? - h)
2
+ ( ?? - ?? )
2
= ?? 2
 are: ?? = h + ???????? ?? ; ?? = ?? +
???????? ?? ; -?? < ?? = ?? where ( h, ?? ) is the centre, ?? is the radius &?? is a parameter. 
Page 5


CIRCLE 
Imagine you have a single point in a wide, empty space. This point is the center of a 
circle. A circle is formed by drawing a line that keeps the same distance all around from 
this center point. Mathematicians love circles because they're simple yet full of surprises. 
In math, a circle is all the points in a plane that are the same distance from a fixed point, 
the center. This fixed distance is called the radius. Circles are cool because they can be 
described in many ways using math equations. Let's explore these equations and look at 
some interesting problems that show off circles' special properties. 
Equation of a circle in various forms: 
(a) The circle with center as origin & radius ' ?? ' has the equation; ?? 2
+ ?? 2
= ?? 2
. 
 
(b) The circle with centre ( h, ?? ) & radius ' ?? ' has the equation; ( ?? - h)
2
+ ( ?? - ?? )
2
= ?? 2
 
(c) The general equation of a circle is ?? 2
+ ?? 2
+ 2???? + 2???? + ?? = 0 
with centre as ( -?? , -?? ) & radius = v?? 2
+ ?? 2
- ?? . 
This can be obtained from the equation ( ?? - h)
2
+ ( ?? - ?? )
2
= ?? 2
 
? ?? 2
+ ?? 2
- 2h?? - 2???? + h
2
+ ?? 2
- ?? 2
= 0 
Take  -h = ?? , -?? = ?? and h
2
+ ?? 2
- ?? 2
= ?? 
Condition to define circle :- 
?? 2
+ ?? 2
- ?? > 0 ? real circle. 
?? 2
+ ?? 2
- ?? = 0 ? point circle. 
?? 2
+ ?? 2
- ?? < 0 ? imaginary circle, with real centre, that is ( -?? , -?? ) 
Note : That every second degree equation in ?? &?? , in which coefficient of ?? 2
 is equal to 
coefficient of ?? 2
 & the coefficient of ???? is zero, always represents a circle. 
(d) The equation of circle with ( ?? 1
, ?? 1
) &( ?? 2
, ?? 2
) as extremeties of its diameter is: 
( ?? - ?? 1
) ( ?? - ?? 2
)+ ( ?? - ?? 1
) ( ?? - ?? 2
) = 0. 
 
This is obtained by the fact that angle in a semicircle is a right angle. 
?  (Slope of PA) (Slope of PB) = -1 
?
?? - ?? 1
?? - ?? 1
·
?? - ?? 2
?? - ?? 2
= -1 ? ( ?? - ?? 1
) ( ?? - ?? 2
)+ ( ?? - ?? 1
) ( ?? - ?? 2
) = 0 
Note that this will be the circle of least radius passing through ( ?? 1
, ?? 1
) &( ?? 2
, ?? 2
) . 
Problem 1 Find the equation of the circle whose centre is ( 0,3) and radius is 3 . 
Solution. The equation of the circle is ( ?? - 0)
2
+ ( ?? - 3)
2
= 3
2
 
? ?? 2
+ ?? 2
- 6?? = 0 
Problem 2 Find the equation of the circle which passes through ( 1, -1) and two of its 
dimeter are ?? + 2?? - 5 = 0 and ?? - ?? + 1 = 0 
Solution. Let ?? be the point of intersection of the lines 
?? + 2?? - 5 = 0 
and ?? - ?? + 1 = 0 
Solving (i) and (ii), we get ?? = 1, ?? = 2. So, coordinates of centre are ( 1,2) . since circle 
passes through ( 1, -1) so 
radius = v( 1 - 1)
2
+ ( 2 + 1)
2
 ?  radius = 3 
Hence the equation of the required circle is ( ?? - 1)
2
+ ( ?? + 2)
2
= 9 
Problem 3 If the equation ?? ?? 2
+ ( ?? - 3) ???? + 3?? 2
+ 6???? + 2???? - 3 = 0 represents the 
equation of a circle then find ?? , ?? 
Solution.  ?? ?? 2
+ ( ?? - 3) ???? + 3?? 2
+ 6???? + 2???? - 3 = 0 
above equation will represent a circle if 
coefficient of ?? 2
= coefficient of ?? 2
 
?? = 3 
coefficient of ???? = 0 
?? = 3 
Problem 4 Find the equation of a circle whose diametric end points are ( ?? 1
, ?? 1
) and 
( ?? 2
, ?? 2
) where ?? 1
, ?? 2
 are the roots of ?? 2
- ???? + ?? = 0 and ?? 1
, ?? 2
 are the roots of ?? 2
- ???? +
?? = 0. 
Solution. We know that the equation of the circle described on the line segment joining 
( ?? 1
, ?? 1
) and ( ?? 2
, ?? 2
) as a diameter is ( ?? - ?? 1
) ( ?? - ?? 2
)+ ( ?? - ?? 1
) ( ?? - ?? 2
) = 0. 
?? 2
+ ?? 2
- ( ?? 1
+ ?? 2
) ?? - ( ?? 1
+ ?? 2
) ?? + ?? 1
?? 2
+ ?? 1
?? 2
= 0 
Here, 
 ?? 1
+ ?? 2
= ?? , ?? 1
?? 2
= ??  ?? 1
+ ?? 2
= ?? , ?? 1
?? 2
= ??  
So, the equation of the required circle is 
?? 2
+ ?? 2
- ???? - ???? + ?? + ?? = 0 
Intercepts made by a circle on the axes: 
Intercepts are the points where the axes meet the circle. 
The intercepts made by the circle ?? 2
+ ?? 2
+ 2???? + 2???? + ?? = 0 on the co-ordinate axes 
are 2v?? 2
- ?? (on ?? -axis) &2v?? 2
- ?? (on ?? -axis) respectively. 
If, 
?? 2
> ?? ?  ???????????? ???????? ?? h?? ?? ???????? ???? ?????? ???????????????? ???????????? .  ?? 2
= ?? 
?  ???????????? ???????? h???? ?? h?? ?? - ???????? .  ?? 2
< ?? 
?  ???????????? ???????? ???????????????????? ?????????? ???? ?????????? ?? h?? ?? - ???????? .   
 
???? = 2???? = 2
v
?? 2
- ?? ?? 2
= 2v?? 2
- ?? 2
= 2v?? 2
+ ?? 2
- ?? - ?? 2
= 2v?? 2
- ?? 
Problem 5 Find the locus of the centre of the circle whose ?? and ?? intercepts are ?? and ?? 
respectively. 
Solution. Equation of circle is ?? 2
+ ?? 2
+ 2???? + 2???? + ?? = 0 ?? intercept = ?? 
2v?? 2
- ?? = ?? ?? 2
- ?? =
?? 2
4
 ?? ?????????????????? = ??  2v?? 2
- ?? = ?? ?? 2
- ?? =
?? 2
4
 #( ???? )  
subtracting equation (i) and (ii) 
?? 2
- ?? 2
=
?? 2
- ?? 2
4
 
hence locus of centre is ?? 2
- ?? 2
=
?? 2
-?? 2
4
 
Parametric equations of a circle: 
The parametric equations of ( ?? - h)
2
+ ( ?? - ?? )
2
= ?? 2
 are: ?? = h + ???????? ?? ; ?? = ?? +
???????? ?? ; -?? < ?? = ?? where ( h, ?? ) is the centre, ?? is the radius &?? is a parameter. 
 
Problem 6 Find the parametric equations of the circle ?? 2
+ ?? 2
+ 4?? + 6?? + 9 = 0 
Solution. We have : ?? 2
+ ?? 2
+ 4?? + 6?? + 9 = 0 
? ( ?? + 2)
2
+ ( ?? + 3)
2
= 2
2
 
So, the parametric equations of this circle are 
?? = -2 + 2?????? ?? , ?? = -3 + 2?????? ?? 
Problem 7 Find the equation of the following curve in cartesian form 
?? + ?? = ?????? ?? , ?? - ?? = ?????? ?? where ?? is the parameter. 
Solution. We have : ?? + ?? = ?????? ??   (i) 
?? - ?? = ?????? ?? #( ???? )  
?   ( ?? )
2
+ ( ???? )
2
   ( ?? + ?? )
2
+ ( ?? - ?? )
2
= 1  ?? 2
+ ?? 2
=
1
2
  
Clearly, it is a circle with centre at ( 0,0) and radius 
1
v 2
. 
Position of a point with respect to a 
circle: