Question

Circles in general Show that the polar equationr² - 2r r₀ cos(θ - θ₀) = R² - r₀²describes a circle of radius R whose center has polar coordinates (r₀, θ₀)

Accepted Answer

Recall the relationship between polar and Cartesian coordinates: $$x = r \cos	heta$$ and $$y = r \sin	heta. $$Similarly, the center of the circle has Cartesian coordinates $$x_0$$ = $$r_0$$ \cos\$$theta_0$$ and $$y_0$$ = $$r_0$$ \sin\$$theta_0. $$Rewrite the given polar equation $$r^2 - 2 r$$ $$r_0$$ \cos(	heta - \$$theta_0$$) = $$R^2$$ - $$r_0^2$ by expressing $$\cos(	heta - \$$theta_0$$)$$ using the cosine difference identity: $$\cos(	heta - \$$theta_0$$) = \cos	heta \cos\$$theta_0$$ + \sin	heta \sin\$$theta_0. $$Substitute the expressions for $$\cos	heta$$ and $$\sin	heta in $$terms of $$x$$ and $$y$$, and similarly for $$\cos\$$theta_0$$ and $$\sin\$$theta_0$$, to rewrite the equation entirely in terms of $$x$$, $$y$$, $$x_0$$, and $$y_0. $$Simplify the equation to get it into the standard form of a circle: $$(x - x_0$)^2 + (y$$ - $$y_0)^2$ = R^2$. This will involve expanding and rearranging terms to isolate the squared terms of (x$$ - $$x_0$$)$$ and (y$$ - $$y_0$$)$$. Conclude that since the equation matches the standard form of a circle with center $$($$x_0$$, $$y_0$$)$$ and radius R$, the original polar equation indeed describes a circle with center at polar coordinates $$($$r_0$$, \$$theta_0$$)$$ and radius R$.

Circles in general Show that the polar equation
r² - 2r r₀ cos(θ - θ₀) = R² - r₀²
describes a circle of radius R whose center has polar coordinates (r₀, θ₀)

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Key Concepts

Polar Coordinates and Conversion to Cartesian Coordinates

Equation of a Circle in Cartesian Coordinates

Use of the Law of Cosines in Polar Form