Understanding Engineering Mathematics

The second equation is actually the same as the first – just cancel 2

throughout the equation. So these equations actually represent the same

line and they therefore intersect at every point – there is an infinite

number of solutions.

Solution to review question 7.1.6

You will save time by noting that the lines (i)x+y=3 and (ii) 2x+ 2 y=

−1 are in fact parallel lines – they have the same gradient,−1. So they

never intersect, and we have only to consider the intersection of each of

them with the line (iii)y= 3 x−1.

(i) and (iii) intersect where

y= 3 −x= 3 x− 1

Solving forxand thenygives

x= 1 ,y= 2

(ii) and (iii) intersect where

y=−^12 −x= 3 x− 1

or

x=^18 ,y=−^58

7.2.7 Equation of a circle

➤

205 222➤

Two things characterise a circle uniquely – its centre and its radius. Suppose therefore
we refer a circle to a set of Cartesian axes in which its centre has coordinates (a,b)and
its radius isr. Then we know that any point (x,y) on the circle is always a distancer
from the centre (a,b). So, using Pythagoras’ theorem (154
➤
yet again) we can write
the distance from (x,y) to the centre (a,b)as

√
(x−a)^2 +(y−b)^2. Since this is always
equal to the radiusrwe can then write

(x−a)^2 +(y−b)^2 =r^2

This, with the square root sign removed by squaring, is the equation orlocusfor a circle
with centre (a,b) and radiusr, in Cartesian coordinates. This equation may be expanded
(42
➤
) to give the form

x^2 +y^2 + 2 fx+ 2 gy+h= 0

Problem

Expressf,g,hin terms ofa,b,r.

You should find

f=−a, g=−b, h=a^2 +b^2 −r^2