C2 Circles and Completing the Square

I like to revise by writing things down or typing them up (helps me remember) so voila.

These are examples of simple completing the square and using completing the square in the equation of a circle

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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

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Completing the square (cont.)

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

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Equation of a Circle

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)

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Completing the square and the equation of a circle

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)

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Taking a factor

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)

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