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what do u think an expression for r is? in terms of x and y
How did you graph that? o.0
u click on draw button
em, well I tried multiplying both sides by r and got r^2=8rcos(theta) (x^2+y^2)=8x
yep that is right
haha i know that. But i meant how did you come up with that graph.
but i think it needs to be simplified further than that. right?
thats the relationship between polar and cartesian
(x^2+y^2)-8x = 0 now what?
u can put it in circle form
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
I don't know how to complete squares :/ that is where I got stuck.
watch youtube video, its faster
its going to look like this
(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
hrm ok. I tried looking for a good youtube video but just got all confused =.=
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.
um completing square therse nothing to it
haha sigh, i know it should be super easy. But not for me xD
Any quadratic is in the form ax^2+bx+c, where a b and c are anything
okay no lets change this... to something simpler first
x^2+bx+c lets say u have an equation of this form
to put in square form, okay so u need a bracket with the variable + a constant squares
(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
x^2+bx+c ---> becomes (x+b/2)^2-(b/2)^2+c
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
k..i'm trying to follow hah
and then we just put the y^2 back in there?