Graph and convert this equation to rectangular coordinates
polar equation: r=8cos(theta)

- Babynini

Graph and convert this equation to rectangular coordinates
polar equation: r=8cos(theta)

- schrodinger

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- dan815

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- dan815

|dw:1434650153992:dw|

- dan815

what do u think an expression for r is? in terms of x and y

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## More answers

- Babynini

How did you graph that? o.0

- dan815

u click on draw button

- Babynini

em, well I tried multiplying both sides by r and got
r^2=8rcos(theta)
(x^2+y^2)=8x

- dan815

yep that is right

- Babynini

haha i know that. But i meant how did you come up with that graph.

- Babynini

but i think it needs to be simplified further than that. right?

- dan815

thats the relationship between polar and cartesian

- Babynini

(x^2+y^2)-8x = 0
now what?

- dan815

u can put it in circle form

- dan815

x^2+y^2=8x
rewrite it as this first
x^2-8x + y^2=0
^------------ complete the square and put the constant on the right side then

- Babynini

I don't know how to complete squares :/ that is where I got stuck.

- dan815

um

- dan815

watch youtube video, its faster

- dan815

its going to look like this

- dan815

(x-4)^2-16+y^2=0
(x-4)^2+y^2=16
(x-4)^2+y^2=4^2
radius 4, shift in x by 4

- dan815

|dw:1434650531548:dw|

- Babynini

hrm ok. I tried looking for a good youtube video but just got all confused =.=

- imqwerty

we have the basic formulas -
x = rcos(theta)
y = rsin(theta)
If x^2 + y^2 = r^2 and r^2 = 8rcos(theta) = 8(x)
then you have been given that: x^2 + y^2 = 8x x^2 + 8x + y^2 = 0
Convert to the center-vertex form for circles.

- dan815

um completing square therse nothing to it

- Babynini

haha sigh, i know it should be super easy. But not for me xD

- dan815

Any quadratic is in the form
ax^2+bx+c,
where a b and c are anything

- dan815

okay no lets change this... to something simpler first

- Babynini

k

- dan815

x^2+bx+c lets say u have an equation of this form

- dan815

to put in square form, okay so u need a bracket with the variable + a constant squares

- dan815

(x+d)^2
okay when u square this what happens
(x+d)(x+d)=x*(x+d) +d*(x+d) = x^2+dx+dx+d^2 = x^2+2dx+d^2
therefore
if you want to rewrite
x^2+bx+c in (x+d)^2 form
then you know that the 2dx has add up to bx
2dx=bx
d=b/2
so we see that it has to be in this form
(x+b/2)^2
now the constant part will be d^2 or
(b/2)^2, this we subtract away as it is in cluded when u square it

- dan815

x^2+bx+c
--->
becomes
(x+b/2)^2-(b/2)^2+c

- dan815

in your case
we have
the equation
x^2+bx+c
x^2-8x+0
which means b=-8 and c=0
okay so
(x+b/2)^2 -(b/2)^2+c
(x+(-8/2))^2 - (-8/2)^2 +0

- Babynini

k..i'm trying to follow hah

- Babynini

and then we just put the y^2 back in there?

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