Question

r csc theta = 3

Answer 1

So far, I have gotten as far as
\[x^2 + y^2 = 3 r \sin \theta\]

Answer 2

\[\sin \theta\] is equivalent to \[\frac{ y }{ r }\] therefore I will set it as \[x^2 + y^2 = 3r (\frac{ y }{ r })\]

Is this correct?

Answer 3

I then cancel out the r, leaving only \[x^2 + y^2 = 3y\]

Answer 4

Would it then be, \[x^2 + y^2 - 3y = 0\]

Answer 5

Afterwards, would I bring over \[x^2\] to make \[y^2 - 3y = -x^2\]

Afterwards I solve for y by plugging in 0 for x.

Answer 6

x^2+y^2=3y is correct

Answer 7

what else do you want to do it with besides covert it to Cartesian form?

Answer 8

I wanted to turn it into rectangular coordinates. I have the solution but I want to know how to get to the solution.

Answer 9

the solution is \[x^2 + (y - \frac{ 3 }{ 2 })^2 = (\frac{ 3 }{ 2 })^2\]

Answer 10

you mean rectangular form (aka cartesian form)
you already have it in rectangular form
x^2+y^2=3y

do you want to be it in the standard from for a circle
(x-h)^2+(y-k)^2=r^2?

Answer 11

just so you know you could have stopped at x^2+y^2=3y since it says write rcsc(theta)=3 in Cartesian form (Aka rectangular form) but if you want to match the book you just complete the square
\[x^2+y^2-3y=0 \\ x^2+y^2-3y+(\frac{3}{2})^2=(\frac{3}{2})^2 \\ x^2+(y-\frac{3}{2})^2=(\frac{3}{2})^2 \]

Answer 12

recall \[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \]

Answer 13

Oh so I do not get rid of x^2?

That makes sense.

Answer 14

no you don't get rid of x^2

Answer 15

Okay I should be good for this then. Thank you very much for your help :)

Answer 16

with*

Sorry my grammar is a bit off.

Answer 17

here a another example putting a circle in center radius form:

\[x^2+y^2+6x-5y=0 \\ x^2+6x+y^2-5y=0 \\ x^2+6x+(\frac{6}{2})^2+y^2-5y+(\frac{5}{2})^2=(\frac{6}{2})^2+(\frac{5}{2})^2 \\ (x+\frac{6}{2})^2+(y-\frac{5}{2})^2=\frac{36}{4}+\frac{25}{4} \\ (x+3)^2+(y-\frac{5}{2})^2=\frac{91}{4}\]
\[(x+3)^2+(y-\frac{5}{2})^2=(\frac{\sqrt{91}}{2})^2 \]

Answer 18

you have to complete the square for both the x group of terms and the y group of terms

Answer 19

and whatver you add to one side you add to the other

Answer 20

oops I don't know how to add 36+25=61

Answer 21

91 should be 61 :p

Answer 22

Okay that's good to know. However pertaining to my question specifically, I only complete the square for the y variable, correct?

And it's no problem, mistakes happen!

Answer 23

well you don't have any terms that just have x
the x group is already written as a square
that is you just have x^2 no other terms with any kind of x

Answer 24

Alright thank you!
Closing question.