Basic Engineering Mathematics, Fifth Edition

The circle 237

x^2 +y^2 + 8 x− 2 y+ 8 =0 is of the form shown in
equation (2),

where a=−

(

8

2

)

=− 4 ,b=−

(

− 2

2

)

= 1

and r=

√

(− 4 )^2 + 12 − 8 =

√

9 = 3.

Hence,x^2 +y^2 + 8 x− 2 y+ 8 =0 represents a circle
centre(− 4 , 1 )andradius 3, as shown in Figure 26.17.

a= 4

b=1

 2

2

4

y

 80  6  4

r^ =

3

x

Figure 26.17

Alternatively,x^2 +y^2 + 8 x− 2 y+ 8 =0 may be rear-
ranged as

(x+ 4 )^2 +(y− 1 )^2 − 9 = 0

i.e. (x+ 4 )^2 +(y− 1 )^2 = 32

which represents a circle,centre(− 4 , 1 )andradius 3,
as stated above.

Problem 19. Sketch the circle given by the

equationx^2 +y^2 − 4 x+ 6 y− 3 = 0

The equation of a circle, centre(a,b),radiusr is
given by

(x−a)^2 +(y−b)^2 =r^2

The general equation of a circle is
x^2 +y^2 + 2 ex+ 2 fy+c= 0

From abovea=−

2 e

2

,b=−

2 f

2

andr=

√

a^2 +b^2 −c

Hence, ifx^2 +y^2 − 4 x+ 6 y− 3 = 0

then a=−

(

− 4

2

)

= 2 ,b=−

(

6

2

)

=−3and

r=

√

22 +(− 3 )^2 −(− 3 )=

√

16 = 4

Thus, the circle hascentre( 2 ,− 3 )andradius 4,as

shown in Figure 26.18.

 4

 2

2

3

4

1

y

 4

 5

 7

 8

 3

 2 0 2 4 6 x

r^ =^

4

Figure 26.18

Alternatively,x^2 +y^2 − 4 x+ 6 y− 3 =0 may be rear-

ranged as

(x− 2 )^2 +(y+ 3 )^2 − 3 − 13 = 0

i.e. (x− 2 )^2 +(y+ 3 )^2 = 42

which represents a circle,centre( 2 ,− 3 )andradius 4,

as stated above.

Now try the following Practice Exercise

PracticeExercise 104 The equation of a

circle (answers on page 351)

1. Determine (a) the radius and (b) the
   co-ordinates of the centre of the circle given
   by the equationx^2 +y^2 − 6 x+ 8 y+ 21 =0.


2. Sketch the circle given by the equation
   x^2 +y^2 − 6 x+ 4 y− 3 =0.


3. Sketch the curvex^2 +(y− 1 )^2 − 25 =0.


4. Sketch the curvex= 6

√[

1 −

(y

6

) 2 ]