How do you convert #4=(x-4)^2+(y-1)^2# into polar form?

Question

James Adkins · Accepted Answer

# r^2 - 8r \cos heta - 2 r \sin heta + 13 = 0# There is a standard way to convert from rectangular coordinates to polar coordinates: replace all instances of #x# with #r\cos heta# and all instances of #y# with #r \sin heta#. The equation of the curve (a circle) can be expanded and written as follows: #x^2 + y^2 - 8x - 2y + 13 = 0# Now do all the replacement work. #(r \cos heta)^2 + (r \sin heta)^2 - 8 r \cos heta - 2 r \ sin heta + 13 = 0# #\Rightarrow r^2 - 8 r \cos heta - 2 r \sin heta + 13 = 0#

Kai Anderson · Answer

undefined To convert the equation $4 = (x - 4)^2 + (y - 1)^2$ into polar form, we use the following relationships between Cartesian and polar coordinates:

$$ x = r \cos(\theta) $$
$$ y = r \sin(\theta) $$

Substitute these expressions into the given equation and simplify:

$$ 4 = (r \cos(\theta) - 4)^2 + (r \sin(\theta) - 1)^2 $$

$$ 4 = (r^2 \cos^2(\theta) - 8r\cos(\theta) + 16) + (r^2 \sin^2(\theta) - 2r\sin(\theta) + 1) $$

$$ 4 = r^2 \cos^2(\theta) + r^2 \sin^2(\theta) - 8r\cos(\theta) - 2r\sin(\theta) + 17 $$

Using the trigonometric identity $ \cos^2(\theta) + \sin^2(\theta) = 1 $, the equation simplifies to:

$$ 4 = r^2 - 8r\cos(\theta) - 2r\sin(\theta) + 17 $$

$$ 0 = r^2 - 8r\cos(\theta) - 2r\sin(\theta) + 13 $$

This is the polar form of the equation.