Question

find the center and radius of the circle x^2+y^2+8x divided by 6y =0

Answer 1

"x^2+y^2+8x divided by 6y =0 "
you sure that's a circle?

Answer 2

I know it looks werid thats why I am stuck but yes I am sure its suppossed to be a circle @dpaInc

Answer 3

hmm... i guess it is a circle.. a partial one...
http://www.wolframalpha.com/input/?i=graph+%28x%5E2%2By%5E2%2B8x%29%2F%286y%29+%3D0

Answer 4

wow that sites awesome! but that still dosent answer my question... @dpaInc

Answer 5

do you know the "general form" for the equation of a circle?

Answer 6

ya is that x^2+Y^2=R^2

Answer 7

not quite - that is for a circle centered at the origin. what about a circle with center at (a,b) and radius r?

Answer 8

ok then its (x-x)^2+(Y-Y)^2=r^2

Answer 9

almost - I think you meant:\[(x-a)^2+(y-b)^2=r^2\]

Answer 10

ya I did but how does that help...

Answer 11

we'll get there - patience....

now, the equation you are given is as follows:\[\frac{x^2+y^2+8x}{6y}=0\]correct?

Answer 12

yes well i have it written as \[x ^{2}+y ^{2}+8x \div 6y=0\]but yes

Answer 13

I wanted to confirm that the entire equation is divided by 6y and not just the 8x part.

Answer 14

no i dont think the whole equation is divided by 6

Answer 15

so is it actually like this:\[x^2+y^2+\frac{8x}{6y}=0\]

Answer 16

I think so it is written like I put above^^^^

Answer 17

then it cannot be a circle - look here: http://www.wolframalpha.com/input/?i=graph+%28x^2%2By^2%29%2B%288x%29%2F%286y%29+%3D0

I am sure it must be of the form:\[\frac{x^2+y^2+8x}{6y}=0\]

Answer 18

ok then you are correct

Answer 19

ok, so now multiply both sides of your equation by 6y - what will you end up with?

Answer 20

\[x ^{2}+y ^{2}+8y=6y\]....

Answer 21

on the left-hand-side you seem to have changed the 8x to 8y and on the right-hand-side \(0\times 6y\ne6y\)

Answer 22

sorry i ment 8x and of course the left side is still zero sorry....

Answer 23

ok, so what we actually end up with is:\[x^2+y^2+8x=0\]agreed?

Answer 24

yes

Answer 25

now what you need to do next is to transform \(x^2+8x\) into a form more like \((x+a)^2\). but:\[(x+a)^2=x^2+2ax+a^2\]so we need to subtract \(a^2\) in order to be left with just \(x^2+2ax\).
therefore:\[x^2+8x=(x+?)^2-?\]can you work out what the question marks should be?

Answer 26

wait what does the a represent..

Answer 27

'a' is just some constant number

Answer 28

look at it this way, we have shown that:\[x^2+2ax=(x+a)^2-a^2\]and we need to find:\[x^2+8x=?\]

Answer 29

so what value should we set 'a' equal to?

Answer 30

is \[a\] the same as \[y\] in the last equation

Answer 31

no

Answer 32

e.q. if we set a=5 we will get:\[x^2+10x=(x+5)^2-25\]

Answer 33

! am sorry .....but i dont understand why you indrouced a new variable

Answer 34

ok, let me try another approach...

Answer 35

we ended up with this:\[x^2+y^2+8x=0\]agreed?

Answer 36

yes thats the last thing i understood

Answer 37

and we can rearrange this to get:\[x^2+8x+y^2=0\]

Answer 38

now we want to try and get the terms \(x^2+8x\) into a form that resembles the general equation of a circle. this is typically called completing the square.

Answer 39

do you agree that:\[(x+n)^2=x^2+2nx+n^2\]where 'n' is just some number?

Answer 40

OHHHH I get it now....

Answer 41

good :)

so we can rearrange this expansion to get:\[x^2+2nx=(x+n)^2-n^2\]

Answer 42

and what we are trying to do is to get \(x^2+8x\) into such a form

Answer 43

ya i do and would n be the second term so 8x divided by 2 squared making it 16

Answer 44

8 divided by 2 is not 16 :)

Answer 45

no 8 divided by 8 = 4 then 4^2 is 16

Answer 46

compare these two:\[x^2+2nx\]\[x^2+8x\]what should n equal to make them equal?

Answer 47

4

Answer 48

correct, so we now know we can write:\[x^2+8x=(x+4)^2-4^2=(x+4)^2-16\]agreed?

Answer 49

yes and then slove for x

Answer 50

no

Answer 51

what we have just shown is just another way of writing \(x^2+8x\) - nothing to solve here

Answer 52

ok....

Answer 53

so, going back to our last equation, we had:\[x^2+8x+y^2=0\]we can now replace \(x^2+8x\) with our alternative to get:\[(x+4)^2-16+y^2=0\]make sense so far?

Answer 54

yes that does

Answer 55

now we rearrange to move the -16 to the right-hand-side of the equals sign to get:\[(x+4)^2+y^2=16\]which is almost in the final circle form

Answer 56

remember the general equation of a circle is given by:\[(x-a)^2+(y-b)^2=r^2\]where the center is located at (a,b) and the radius is r.

Answer 57

so we can now rearrange once more to get:\[(x-(-4))^2+(y-0)^2=4^2\]

Answer 58

does that make sense?

Answer 59

yes it does i think i can work out the rest i have to go now sorry but you are A GOD THANK YOU SO MUCH!!!!!!!!

Answer 60

yw :)

Answer 61

BTW: this is what your original equation looks like: http://www.wolframalpha.com/input/?i=graph+%28x^2%2By^2%2B8x%29%2F%286y%29+%3D0+for+x%3D-8+to+0+and+y%3D-4+to+4

Answer 62

notice it has a little line in the center of the circle

Answer 63

this is because your original equation was divided by 6y so this means when y=0 we get division by zero - so this point needs to be excluded from the circle.