Understanding Engineering Mathematics

Often, it is the latter form of the equation of a circle that is given, and you are asked to
find the centre and radius. To do this we complete the square (66
➤
)onx,yand return
to the first form, from which centre and radius can be read off directly.
Note that in polar coordinates, the equation of a circle with radiusaand centre at the
origin takes the very simple formr=a. This illustrates how useful polar coordinates can
be in some circumstances.

Solution to review question 7.1.7

A.The equation of the circle centre (a,b) with radiusris

(x−a)^2 +(y−b)^2 =r^2

Applying this to the data given produces:

(i) For centre (0, 0) and radius 2 we get

(x− 0 )^2 +(y− 0 )^2 = 22 , orx^2 +y^2 = 4

(ii) For centre (1, 2) radius 1 we get

(x− 1 )^2 +(y− 2 )^2 =1orx^2 +y^2 − 2 x− 4 y+ 4 = 0

(iii) For centre (−1, 4) and radius 3 we get

(x+ 1 )^2 +(y− 4 )^2 =9orx^2 +y^2 + 2 x− 8 y+ 8 = 0

B. To find the centre and radius of a circle whose equation is given in the

formx^2 +y^2 + 2 fx+ 2 gy+h=0 we re-express it in the form

(x−a)^2 +(y−b)^2 =r^2

by completing the square inxandy(66

➤

).

(i)x^2 +y^2 − 2 x+ 6 y+ 6 =x^2 − 2 x+y^2 + 6 y+ 6

=(x− 1 )^2 − 1 +(y+ 3 )^2 − 9 + 6

on completing the square forx^2 − 2 xandy^2 + 6 y

=(x− 1 )^2 +(y+ 3 )^2 − 4

So the equation is equivalent to

(x− 1 )^2 +(y+ 3 )^2 = 4 = 22

which gives a centre of (1,−3) and radius 2.

(ii) x^2 +y^2 + 4 x+ 4 y− 1 =x^2 + 4 x+y^2 + 4 y− 1

=(x+ 2 )^2 − 4 +(y+ 2 )^2 − 4 − 1

=(x+ 2 )^2 +(y+ 2 )^2 − 9

on completing the square. So the equation becomes

(x+ 2 )^2 +(y+ 2 )^2 = 9 = 32

giving a centre (− 2 ,−2) and a radius 3.