What Are the Boundaries of θ and r for a Circle Centered at (1,2)?

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SUMMARY

The discussion centers on determining the boundaries of the polar coordinates θ and r for the circle defined by the equation (x - 1)² + (y - 2)² = 1. The radius r is established to be between 0 and 1, while the angle θ ranges from 0 to π/2, indicating that the circle is situated in the first quadrant and touches both the x-axis and y-axis. Additionally, when redefining the equations as y = 2 + rsinθ and x = 1 + rcosθ, the circle remains defined within the same boundaries for r, but the angle θ can extend from 0 to 2π.

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Homework Statement


We have the circle [itex](x - 1)^2 + (y-2)^2 = 1[/itex]. Find the boundaries of θ and r.

Homework Equations


x = h + rcosθ
y = k + rsinθ

The Attempt at a Solution


This is a circle with its origin at (1,2) and a radius of 1 so r is between 0 and 1 and the circle lies in the first quadrant but touches the x and y-axis so theta is between 0 and pi/2

Is this correct? Just curious

edit: additional question: let's say I define y = 2 + rsinθ and x = 1 + rcosθ. Would this circle go from 0 to 2pi? (point of origin (1,2))
 
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Jonmundsson said:

Homework Statement


We have the circle [itex](x - 1)^2 + (y-2)^2 = 1[/itex]. Find the boundaries of θ and r.


Homework Equations


x = h + rcosθ
y = k + rsinθ

The Attempt at a Solution


This is a circle with its origin at (1,2) and a radius of 1 so r is between 0 and 1 and the circle lies in the first quadrant but touches the x and y-axis so theta is between 0 and pi/2

Is this correct? Just curious

edit: additional question: let's say I define y = 2 + rsinθ and x = 1 + rcosθ. Would this circle go from 0 to 2pi? (point of origin (1,2))

Your comment in the edit is exactly how to set up the problem. You would put ##r = 1## to get the circle and ##0 \le r \le 1## to get the interior.
 

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