how would you find the polar equation of a circle if you're given the circle's center and radius?

Answer 1

In polar coordinates, equation of a circle at with its origin at the center is simply:

r² = R²

A circle, with C(ro,to) as center and R as radius, has has a polar equation:

r² - 2 r ro cos(t - to) + ro² = R²

Notice how this becomes the same as the first equation when ro = 0, to = 0

Answer 2

Polar Equation Of A Circle

Answer 3

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RE:
how would you find the polar equation of a circle if you're given the circle's center and radius?

Answer 4

In rectangular coordinates the circle would be:
(x-a)² + (y-b)² = R²
changing this to polar would give:
(rsinθ-a)² + (rcosθ-b)² = R²
rsin²θ - 2arsinθ + a² + r²cos²θ - 2brcosθ + b² = R²
r² - r(2asinθ + 2bcosθ) + a² + b² - R² = 0
Using the quadratic equation:
r = bcosθ+asinθ±√[(bcosθ+asinθ)²-a²-b²+R²]

Answer 5

the formula for a circle is x^2+y^2=a^2 and to convert cartesian to polar use x=rcos(theta) and y=rsin(theta)
therefore the equation becomes:
(r^2) (sin(theta))^2+(r^2) (cos(theta))^2 = a^2
simplify: factor out the r^2 and (sin(x))^2+(cos(x))^2=1 therefore
(r^2) (sin(theta))^2+(r^2) (cos(theta))^2) = a^2
r^2(1)=a^2
and finally r=a where a is a constant and the actual radius of the circle

Answer 6

x² + y² = r² . . . . circle is at the origin
(x-h)² + (y-p)² = r². . . . .. center of circle is at . . (h,p)
(r cosA -h)² + (r sinA - p)² = r²

Answer 7

(y-a)+(x-b)=r