Higher Engineering Mathematics, Sixth Edition

128 Higher Engineering Mathematics

5

4

2

024 x

y

r^5

2

b 53

a 52

Figure 13.14

Comparing this with equation (2) gives:

2 e=− 2 a,i.e.a=−

2 e

2

and 2f=− 2 b,i.e.b=−

2 f

2

and c=a^2 +b^2 −r^2 ,

i.e., r=

√

(a^2 +b^2 −c)

Thus, for example, the equation

x^2 +y^2 − 4 x− 6 y+ 9 = 0

represents a circle with centre a=−

(

− 4

2

)

,

b=−

(

− 6

2

)

,i.e.at(2,3)andradius

r=

√

( 22 + 32 − 9 )=2.

Hencex^2 +y^2 − 4 x− 6 y+ 9 =0isthecircleshownin

Fig. 13.14 (which may be checked by multiplying out

the brackets in the equation

(x− 2 )^2 +(y− 3 )^2 = 4

Problem 16. Determine (a) the radius, and (b) the

co-ordinates of the centre of the circle given by the

equation:x^2 +y^2 + 8 x− 2 y+ 8 =0.

x^2 +y^2 + 8 x− 2 y+ 8 =0isoftheformshowninequa-

tion (2),

wherea=−

(

8

2

)

=− 4 ,b=−

(

− 2

2

)

= 1

and r=

√

[(− 4 )^2 +( 1 )^2 −8]=

√

9 = 3

Hencex^2 +y^2 + 8 x− 2 y+ 8 =0 represents acirclecen-

tre (−4, 1)andradius 3, as shown in Fig. 13.15.

a 524

b 51

22

2

4

y

28026 24

r^5

3

x

Figure 13.15

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 17. Sketch the circle given by the

equation:x^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

and

r=

√

(a^2 +b^2 −c).

Hence ifx^2 +y^2 − 4 x+ 6 y− 3 = 0

then a=−

(

− 4

2

)

= 2 , b=−

(

6

2

)

=− 3

and r=

√

[( 2 )^2 +(− 3 )^2 −(− 3 )]

=

√

16 = 4

Thusthecirclehascentre(2,−3) and radius 4,as

shown in Fig. 13.16.

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