Answer:
Step-by-step explanation:
The identities you need here are:
[tex]r=\sqrt{x^2+y^2}[/tex] and [tex]r^2=x^2+y^2[/tex]
You also need to know that
x = rcosθ and
y = rsinθ
to get this done.
We have
r = 6 sin θ
Let's first multiply both sides by r (you'll always begin these this way; you'll see why in a second):
r² = 6r sin θ
Now let's replace r² with what it's equal to:
x² + y² = 6r sin θ
Now let's replace r sin θ with what it's equal to:
x² + y² = 6y
That looks like the beginnings of a circle. Let's get everything on one side because I have a feeling we will be completing the square on this:
[tex]x^2+y^2-6y=0[/tex]
Complete the square on the y-terms by taking half its linear term, squaring it and adding it to both sides.
The y linear term is 6. Half of 6 is 3, and 3 squared is 9, so we add 9 in on both sides:
[tex]x^2+(y^2-6y+9)=9[/tex]
In the process of completing the square, we created within that set of parenthesis a perfect square binomial:
[tex]x^2+(y-3)^2=9[/tex]
And there's your circle! Third choice down is the one you want.
Fun, huh?
