Finding center of circle with Polar Coordinates

In summary, 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 >.<
  • #1
PsychonautQQ
784
10

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 ;-/
 
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  • #2
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.
 
  • #3
CAF123 said:
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?
 
  • #4
PsychonautQQ said:
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∅.
 
  • #5
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.
 
  • #6
CAF123 said:
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
 
  • #7
eumyang said:
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
 
  • #8
PsychonautQQ said:
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
 
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  • #9
PsychonautQQ said:
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)?
 
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  • #10
PsychonautQQ said:
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.
 
  • #11
eumyang said:
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 -_-
 
  • #12
PsychonautQQ said:
(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
 
  • #13
ehild said:
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?
 
  • #14
PsychonautQQ said:
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.
 
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  • #15
CAF123 said:
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?
 
  • #16
PsychonautQQ said:
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
 
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  • #17
ehild said:
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
 
  • #19
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.
 

1. How do you use polar coordinates to find the center of a circle?

To find the center of a circle using polar coordinates, you need to know the radius of the circle and the angle of the radius. The center of the circle will be at the point where the radius intersects the origin (0,0) on the polar coordinate plane.

2. What is the formula for finding the center of a circle using polar coordinates?

The formula for finding the center of a circle with polar coordinates is (r,θ), where r is the radius of the circle and θ is the angle of the radius in radians.

3. Can you use polar coordinates to find the center of a circle if the radius is not given?

No, you cannot use polar coordinates to find the center of a circle if the radius is not given. Polar coordinates require the radius in order to determine the center of the circle.

4. How do you convert Cartesian coordinates to polar coordinates to find the center of a circle?

To convert Cartesian coordinates (x,y) to polar coordinates (r,θ), you can use the following formulas: r = √(x^2 + y^2) and θ = arctan(y/x). Once you have the polar coordinates, you can find the center of the circle using the aforementioned formula (r,θ).

5. Are there any other methods for finding the center of a circle besides using polar coordinates?

Yes, there are other methods for finding the center of a circle, such as using the distance formula or using the equation of a circle. However, using polar coordinates is a common and effective method for finding the center of a circle.

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