Higher Engineering Mathematics

THE CIRCLE AND ITS PROPERTIES 141

B

For example, Fig. 14.8 shows a circlex^2 +y^2 =9.
More generally, the equation of a circle, centre (a,b),
radiusr, is given by:

(x−a)^2 +(y−b)^2 =r^2 (1)

Figure 14.9 shows a circle (x−2)^2 +(y−3)^2 =4.
The general equation of a circle is:

x^2 +y^2 + 2 ex+ 2 fy+c=0(2)

Figure 14.8

Figure 14.9

Multiplying out the bracketed terms in equation (1)
gives:

x^2 − 2 ax+a^2 +y^2 − 2 by+b^2 =r^2

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) and radius

r=

√

(2^2 + 32 −9)=2.

Hence x^2 +y^2 − 4 x− 6 y+ 9 =0 is the circle

shown in Fig. 14.9 (which may be checked by

multiplying out the brackets in the equation

(x−2)^2 +(y−3)^2 = 4

Problem 12. 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 =0 is of the form shown in

equation (2),

wherea=−

( 8

2

)

=−4,b=−

(− 2

2

)

= 1

and r=

√

[(−4)^2 +(1)^2 −8]=

√

9 = 3

Hence x^2 +y^2 + 8 x− 2 y+ 8 =0 represents a

circlecentre (−4, 1)andradius 3, as shown in

Fig. 14.10.

Alternatively,x^2 +y^2 + 8 x− 2 y+ 8 =0 may be

rearranged 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) and

radius 3, as stated above.

Figure 14.10