128 Higher Engineering Mathematics
5
4
2
0
2
4
x
y
r 5
2
b 5 3
a 5 2
Figure 13.14
Comparing this with equation (2) gives:
2e = −2a, i.e. a = −
2e
2
and 2 f = −2b, i.e. b = −
2f
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
− 4x − 6y + 9 = 0
represents a circle with centre a =−
−4
2
,
b = −
−6
2
, i.e. at (2, 3) and radius
r =
(2 2 + 3 2 − 9) = 2.
Hence x 2 + y 2 − 4x − 6y + 9 =0 is the circle shown in
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 + 8x − 2y + 8 =0.
x 2 + y 2 + 8x − 2y + 8 =0 is of the form shown in equation (2),
where a = −
8
2
= −4, b = −
−2
2
= 1
and r =
[(−4) 2 + (1) 2 − 8] =
√
9 = 3
Hence x 2 + y 2 + 8x − 2y + 8 =0 represents a circle centre (−4, 1) and radius 3, as shown in Fig. 13.15.
a 524
b 51
22
2
4
y
24
26
28
0
r 5
3
x
Figure 13.15
Alternatively, x 2 + y 2 + 8x − 2y + 8 = 0 may be rearranged as:
(x + 4)
2
+ ( y − 1)
2
− 9 = 0
i.e.
( x + 4)
2
+ ( y − 1)
2
= 3
2
which represents a circle, centre (−4, 1) and radius 3,
as stated above.
Problem 17. Sketch the circle given by the
equation: x 2 + y 2 − 4x + 6y − 3 = 0.
The equation of a circle, centre (a, b), radius r is
given by:
(x − a)
2
+ ( y − b)
2
= r
2
The general equation of a circle is
x
2
+ y
2
+ 2ex + 2 f y + c = 0.
From above a =−
2e
2
, b =−
2 f
2
and
r =
(a 2 + b 2 − c).
Hence if x 2 + y 2 − 4x + 6y − 3 =0
then a =−
−4
2
= 2, b =−
6
2
= −3
and r =
[(2) 2 + (−3) 2 − (−3)]
=
√
16 = 4
Thus the circle has centre (2, −3) and radius 4, as
shown in Fig. 13.16.
Alternatively, x 2 + y 2 − 4x + 6y − 3 =0 may be rearranged as:
(x − 2)
2
+ ( y + 3)
2
− 3 − 13 = 0
i.e.
( x − 2)
2
+ ( y + 3)
2
= 4
2