Finding center of circle with Polar Coordinates

[PsychonautQQ]

Homework Statement

r=7sin(∅)
find the center of the circle in Cartesian coordinates and the radius of the circle

The Attempt at a Solution

My math teacher is impossible to understand >.< and then the stupid homework is online and crap blah this class but I REALLY want to understand the material well.. anyway done venting to earn your pity but I REALLY don't know how to do this one and I locked my textbook in my car lol but since I know that isn't acceptable on physics forums I will try >.<

Okay I really don't know.. maybe I shouldn't have started this thread yet... I'll go research online and edit this post soon if anyone wants to be cool and post hints even though I'm breaking the rules that'd be cool.

my attempt:
r=7sin(∅)
x=rcos(∅)
y=rsin(∅)

x^2 + y^2 = r^2

plugging r into the x and y equations and squaring...
49(sin^2(∅)*cos^2(∅)) + 49sin^4(∅) = 49sin^2(∅)
(sin^2(∅)*cos^2(∅)) + sin^4(∅) = sin^2(∅)
cos^2(∅) + sin^2 (∅) = 1
is what that reduces to... alright sooo that's kind of cool I accidently did that but I got no closer to finding the center.. lol ;-/

 

[CAF123]

You know that x²+ y²=r², so you can find what r is. There is also a very simple equation for sin∅ from elementary trig which you can use.

 

[PsychonautQQ]

You know that x²+ y²=r², so you can find what r is. There is also a very simple equation for sin∅ from elementary trig which you can use.

Should I be getting the answer that r = 1?
If so any tips on how I go about finding the center?

 

[CAF123]

Should I be getting the answer that r = 1?
If so any tips on how I go about finding the center?

How did you get r=1?

Make a right angle triangle with base of length x and height y. Let ∅ be the angle between hypotenuse and base. Find sin∅ in terms of x and y. You should also be able to find r, so sub all these expressions into r = 7sin∅.

 

[eumyang]

OP: I mentioned this before, but please stop using the empty set symbol (∅) for theta (θ). Theta can be found in the 3rd row of the "Quick Symbols" box. Sorry to nitpick.

I would like to propose a slightly different way to solve the problem. You've given us the conversion equations:
x = r cos θ
y = r sin θ
x² + y² = r²

Take the original equation
r = 7 sin θ
and multiply both sides by r. Start substituting. You'll need to complete the square at some point.

 

[PsychonautQQ]

How did you get r=1?

Make a right angle triangle with base of length x and height y. Let ∅ be the angle between hypotenuse and base. Find sin∅ in terms of x and y. You should also be able to find r, so sub all these expressions into r = 7sin∅.

sin(θ)= y/r

x=rcos(θ)
x/cos(θ)=r

sin(θ) = y/(x/cos(θ))
sin(θ) = ycos(θ)/x

r=7sin(θ)
r= 7ycos(θ))/x

ahh.. am I on the right track here?
maybe if I backtrack to
sin(θ) = ycos(θ)/x
tan(θ) = y/x

 

[PsychonautQQ]

OP: I mentioned this before, but please stop using the empty set symbol (∅) for theta (θ). Theta can be found in the 3rd row of the "Quick Symbols" box. Sorry to nitpick.

I would like to propose a slightly different way to solve the problem. You've given us the conversion equations:
x = r cos θ
y = r sin θ
x² + y² = r²

Take the original equation
r = 7 sin θ
and multiply both sides by r. Start substituting. You'll need to complete the square at some point.

r^2 = r7sinθ

r=x/cos(θ)
r=y/sin(θ)
x^2/cos^2 = y/sin(θ) * 7sinθ
x^2/cos^2 = 7y

idk what I'm doing

 

[Mark44]

r^2 = r7sinθ

r=x/cos(θ)
r=y/sin(θ)
x^2/cos^2 = y/sin(θ) * 7sinθ
x^2/cos^2 = 7y

idk what I'm doing

Starting from r² = 7rsin(θ), replace r and θ to get the equation in Cartesian form.

You know (I hope!) that:
r² = x² + y²
r sin(θ) = y
r cos(θ) = x

 

[eumyang]

r^2 = r7sinθ

r=x/cos(θ)
r=y/sin(θ)
x^2/cos^2 = y/sin(θ) * 7sinθ
x^2/cos^2 = 7y

idk what I'm doing

I see. I don't want to sound mean, but you need to do some review. And you're thinking too hard. From
r² = 7r sin θ
all you need to do is replace the left side with x² + y² and replace the "r sin θ" on the right side with y, and you get
x² + y² = 7y.
Do you know the equation for a circle with radius r and center at (h, k)?

 

[CAF123]

sin(θ)= y/r

Yes, but you can also express r in terms of x and y. r²= x²+ y² so find r.

x=rcos(θ)
x/cos(θ)=r

What you want to achieve is the polar expression written completely in Cartesian. Writing r=x/cosθ is fine, but this does not eliminate θ, so it is not helpful here.

Your final result should have just x and y.

 

[PsychonautQQ]

I see. I don't want to sound mean, but you need to do some review. And you're thinking too hard. From
r² = 7r sin θ
all you need to do is replace the left side with x² + y² and replace the "r sin θ" on the right side with y, and you get
x² + y² = 7y.
Do you know the equation for a circle with radius r and center at (h, k)?

(x-h)^2 + (y-k)^2 = r^2 ?
I'm having problems solving the formula to make it fit in this form,, too many y's -_-

 

[ehild]

(x-h)^2 + (y-k)^2 = r^2 ?
I'm having problems solving the formula to make it fit in this form,, too many y's -_-

Do not mix r, the polar coordinate, with the radius of the circle.

In Cartesian coordinates, the equation of a circle is ˙(x-h)²+(y-k)²=R².

You already got the equation of the circle in the form x² + y² = 7y which is equivalent with x² -7y+y² = 0. Use the method completing the square.


ehild

 

[PsychonautQQ]

Do not mix r, the polar coordinate, with the radius of the circle.

In Cartesian coordinates, the equation of a circle is ˙(x-h)²+(y-k)²=R².

You already got the equation of the circle in the form x² + y² = 7y which is equivalent with x² -7y+y² = 0. Use the method completing the square. ehild

the x coordinate of the circle is centered at zero?
but how do I complete the square of y^2-7y? y(y-7)? that's not the correct format of (7-k)^2

also you say don't mix the polar coordinate r with the circle radius R, but to get the equation
x^2+y^2 = 7y
I solved for r in
r=7sin(theta)
r^2=r7sin(theta)
r^2=7y (y = rsin(theta)

so I plugged that r into
x^2 + y^2 = R^2, that's not correct?

 

[CAF123]

the x coordinate of the circle is centered at zero?

Yes

but how do I complete the square of y^2-7y? y(y-7)? that's not the correct format of (7-k)^2

Write $y^2 - 7y = (y-b)^2 + c$. Expand the RHS and equate coefficients to solve for b and c.

 

[PsychonautQQ]

Yes

Write $y^2 - 7y = (y-b)^2 + c$. Expand the RHS and equate coefficients to solve for b and c.

-5y = b^2 + c. I don't understand what c is here. I assume b is the y coordinate of the center of the circle. How do I solve for two variables using one equation?

 

[ehild]

the x coordinate of the circle is centered at zero?
but how do I complete the square of y^2-7y? y(y-7)? that's not the correct format of (7-k)^2

also you say don't mix the polar coordinate r with the circle radius R, but to get the equation
x^2+y^2 = 7y
I solved for r in
r=7sin(theta)
r^2=r7sin(theta)
r^2=7y (y = rsin(theta)

so I plugged that r into
x^2 + y^2 = R^2, that's not correct?

No, r is not the same as R. Forget about the polar coordinates. You have the equation for a curve in Cartesian coordinates x,y:

x²+y²-7y=0

How do you know that it is the equation of a circle?

Try to plot it. If y=0 what is x? If x=0, what is y? ehild

 

[PsychonautQQ]

No, r is not the same as R. Forget about the polar coordinates. You have the equation for a curve in Cartesian coordinates x,y:

x²+y²-7y=0

How do you know that it is the equation of a circle?

Try to plot it. If y=0 what is x? If x=0, what is y?


ehild

But to get that I plugged the r from r=7sinθ into it... how did I use that r to get the equation if it's not the same R as x^2 + y^2 = R^2??

Cuz I multipled both sides for r and then plugged in y..
r^2 = 7rsinθ
r^2 = 7y.
x^2 + y^2 = R^2
x^2 + y^2 = 7y
you are okay with doing that to get the equation even though they aren't the same R? I'm so confused

 

[ehild]

Forget about r. Answer: How do you know that x²+y²-7y=0 is equation of a circle?
Read http://www.mathsisfun.com/algebra/circle-equations.html

ehild

 

[PsychonautQQ]

Oh my god you are my savior for posting that link. Thank you everyone in this thread I wouldn't be passing calculus without you.