Find the polar equation of the circle with center on the line theta = pi, of radius 1, and passing through the origin.
The points which correspond to $\theta = \pi$ are exactly those on the $x$-axis. If you want the circle to pass through the origin, the only way to do this is to have it being tangent to $(0,0)$ i.e it has center $(-1,0)$ or $(1,0)$. This equation is given by
$$r(\theta) = (\cos \theta \pm 1, \sin \theta); \ \theta \in [0,2\pi]$$
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Onur Şanlı
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Updated on May 07, 2020Comments
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Onur Şanlı over 3 years
Question
Find the polar equation of the circle with center on the line theta = pi, of radius 1, and passing through the origin.
i set a point (a,pi) on the line and going to this equation
a = rsin(theta) but its wrong.
where did i am doing wrong ?
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Faraad Armwood over 7 yearsI wrote a solution. I understand passing through origin to mean you wanted the circle to pass through the origin. If you meant the line passes through the origin, just take the standard equation of the circle.
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Onur Şanlı over 7 yearsthanks for helping. I understand clearly this question now.
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Faraad Armwood over 7 yearsNo problem and if this is what you are looking for, accept so the question closes.
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amd over 7 yearsWe can’t tell you where you’re going wrong if you don’t show us your work. Please do so.
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amd over 7 yearsThis isn’t a polar equation. It’s a cartesian vector equation with a misleadingly named parameter $\theta$. For a polar equation, you need to express the scalar $r$—the distance from the origin—as a function of the angle $\theta$, which, by the way, will not range from $0$ to $2\pi$.
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Faraad Armwood over 7 years@amd: I know this is a vector parametrization. I simply polarized that x,y coordinates and parametrized the circle the OP wanted. This is what I understood it to mean at the time. However, this is easily manipulated if OP notices $x = \cos \theta +1 , y = \sin \theta$.