C2 Circles and Completing the Square

if given a question asking to complete the square in this format:

x^2 + bx + c 

Then...

x^2 + bx + c = (x + b/2)^2 + c - (b/2)^2

for example:

x^2 + 6x + 3

whic equals...

(x+6/2)^2 + 3 - (6/2)^2  =  (x+3)^2 +3 - 3^2  =  (x+3)^2 + 3 - 9

so the answer is:

(x+3)^2 - 6

dont forget you can test your answer by expanding it

you can also take out factors when completing the square for example:

4x^2 + 16x + 40 = 0

so take the 40 over:

4x^2 + 16x = -40

take out a factor of 4

x^2 + 4x = -10

then complete the square

(x+2)^2 - 4 = -10

(x+2)^2 = - 6

this is primarily used when completing the square to find the equation of a circle as will be demonstrated later

The standard equation for a circle is:

(x-a)^2 + (y-b)^2 = r^2

In this equation:

r is the radius of the circle

a is the x co-ordinate of the centre of the circle

b is the y co-ordinate of the centre of the circle

for exaple:

(x-3)^2 + (y+6) = 9

the radius in 3 (the square root of 9)

and the centre of the circle is at (3,-6)

to find the equation of a circle you might be given a equation looking something like this:

x^2 + y^2 + 24x + 32y + 20 = 0

but its not as hard as it looks! simply complete the square for each unknown, x and y. 

you might want to seperate the two for this and carry c (20 in this case) across so..

(x^2 + 24x) + (y^2 + 32y) = -20

now complete the square

(x+12)^2 -144 + (y+16)^2 - 256 = -20

(x+12)^2 + (y+16)^2 = -20 + 144 + 256

(x+12)^2 + (y+16)^2 = 380

so radius is around 19.5 (bad example I know)

and centre (-12,-16)

As mentioned in card 2 you can take factors when completing the square. You can also do this when finding the equation of a cirlcetike in:

4x^2 + 4y^2 + 16x +24y - 40 = 0

4x^2 + 16x + 4y^2 + 24y = 40

(x^2 + 4x) + (y^2 + 6y) = 10

(x+2)^2 - 4 + (y+3)^2 - 9 = 10

(x+2)^2 + (y+3)^2 = 10 + 4 + 9

(x+2)^2 + (y+3)^2 = 23

radius would be the square root of 23 

and centre (-2,-3)