Question

find an equation equivalent to (x-2)^2 + (y + 2)^2 = 8 in polar coordinates

Answer 1

The circle equation in polar coordinates is:

   $r - 4*cos(\theta) + 4*sin(\theta) = 0$

How to change the coordinates of the circle equation?

Here we have the circle equation:

$(x - 2)^2 + (y + 2)^2 = 8$

First, we expand it to:

$x^2 - 4x + 4 + y^2 + 4y + 4 = 8$

Now we can rewrite it as:

$x^2 + y^2 -4x + 4y + 4 + 4 = 8\\\\x^2 + y^2 - 4x + 4y = 0$

Remember that:

$x^2 + y^2 = r^2\\\\x = r*cos(\theta)\\y = r*sin(\theta)$

Replacing that, we get:

$x^2 + y^2 - 4x + 4y = 0\\\\r^2 - 4r*cos(\theta) + 4r*sin(\theta) = 0$

That is the equation in polar form.

Now, because we can discard the solution r = 0, we can divide both sides by r to get:

$r - 4*cos(\theta) + 4*sin(\theta) = 0$

To simplify it.

If you want to learn more about polar coordinates:

https://brainly.com/question/14965899

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