(x-1)^2+y^2=1 In Polar Form

Jasonbradley

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Jun 17, 2024

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Converting (x-1)^2 + y^2 = 1 to Polar Form

The equation (x-1)^2 + y^2 = 1 represents a circle with center (1, 0) and radius 1. To convert this equation to polar form, we need to use the following relationships:

• x = r cos θ
• y = r sin θ

Let's substitute these into the given equation:

(r cos θ - 1)^2 + (r sin θ)^2 = 1

Expanding the equation:

r^2 cos^2 θ - 2r cos θ + 1 + r^2 sin^2 θ = 1

Simplifying the equation:

r^2 (cos^2 θ + sin^2 θ) - 2r cos θ = 0

Since cos^2 θ + sin^2 θ = 1:

r^2 - 2r cos θ = 0

Factoring out 'r':

r(r - 2 cos θ) = 0

This equation gives us two solutions:

• r = 0
• r = 2 cos θ

The solution r = 0 represents the origin, which is a single point. The solution r = 2 cos θ represents the entire circle.

Therefore, the polar form of the equation (x-1)^2 + y^2 = 1 is r = 2 cos θ.

Understanding the Polar Form

This equation tells us the radius 'r' of any point on the circle is determined by the angle 'θ' it makes with the positive x-axis. As θ changes, the radius 'r' changes accordingly, tracing the shape of the circle.

Here's a breakdown of the equation:

• r = 2 cos θ: This means that the radius 'r' of a point on the circle is twice the value of the cosine of the angle 'θ'.

Key Observations:

• When θ = 0, cos θ = 1, and r = 2. This corresponds to the rightmost point on the circle.
• When θ = π/2, cos θ = 0, and r = 0. This corresponds to the origin.
• When θ = π, cos θ = -1, and r = -2. This corresponds to a point on the circle but is not considered a valid solution in the context of polar coordinates as radius cannot be negative.
• As θ increases from 0 to π, the radius 'r' decreases from 2 to 0, tracing the right half of the circle.
• For θ values between π and 2π, the equation traces the left half of the circle.

This highlights how the polar equation succinctly describes the circle and its properties.