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