Question
Show that the graph of the equation $r=2 a \sin \theta, a>0$, is a circle of radius $a$ with center at $(0, a)$ in rectangular coordinates.
Step 1
The relationships between polar coordinates \((r, \theta)\) and rectangular coordinates \((x, y)\) are given by: \[ x = r \cos \theta \] \[ y = r \sin \theta \] Show more…
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Show that the graph of the equation $r=-2 a \sin \theta, a \geq 0,$ is a circle of radius $a$ with center at $(0,-a)$ in rectangular coordinates.
Show that the graph of the equation r = 2a sin θ, a > 0, is a circle of radius a with center at (0,a) in rectangular coordinates. Transform the equation r = 2a sin θ from polar coordinates to rectangular coordinates. r = 2a sin θ r^2 = 2ar sin θ Multiply each side by a relevant expression. (x^2 + y^2) = 2ay Convert to use rectangular coordinates. x^2 + y^2 - 2ay = 0 Write the equation of a circle in general form. (Do not factor.) x^2 + (y - a)^2 = a^2 Complete the square in y on the left side and factor. Write the equation of a circle in standard form. The polar equation r = 2a sin θ converts to the rectangular equation found above. The rectangular equation is a circle of radius a with center at (0,a).
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