142 GEOMETRY AND TRIGONOMETRY
Problem 13. 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), radiusris
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
Thusthe circle has centre (2,−3) and radius 4,as
shown in Fig. 14.11.
Alternatively,x^2 +y^2 − 4 x+ 6 y− 3 =0 may be
rearranged 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) and
radius 4, as stated above.
− 4 − 2 0 2 4 6 x
4
2
− 2
− 3
− 4
− 8
r=^4
y
Figure 14.11
Now try the following exercise.
Exercise 66 Further problems on the equa-
tion of a circle
- Determine the radius and the co-ordinates of
the centre of the circle given by the equation
x^2 +y^2 + 6 x− 2 y− 26 =0.
[6, (−3, 1)] - Sketch the circle given by the equation
x^2 +y^2 − 6 x+ 4 y− 3 =0.
[Centre at (3,−2), radius 4] - Sketch the curvex^2 +(y−1)^2 − 25 =0.
[Circle, centre (0, 1), radius 5] - Sketch the curvex= 6
√[
1 −(y/ 6 )^2
]
.
[Circle, centre (0, 0), radius 6]
14.6 Linear and angular velocity
Linear velocity
Linear velocityvis defined as the rate of change
of linear displacementswith respect to timet.For
motion in a straight line:
linear velocity=
change of displacement
change of time
i.e. v=
s
t
(1)
The unit of linear velocity is metres per second (m/s).
Angular velocity
The speed of revolution of a wheel or a shaft is
usually measured in revolutions per minute or revo-
lutions per second but these units do not form part
of a coherent system of units. The basis in SI units
is the angle turned through in one second.
Angular velocity is defined as the rate of change of
angular displacementθ, with respect to timet. For an
object rotating about a fixed axis at a constant speed:
angular velocity=
angle turned through
time taken
i.e. ω=
θ
t
(2)