# What are the polar coordinates of (x-1)^2-(y+5)^2=-24?

Dec 17, 2016

Expand the squares, substitute $y = r \sin \left(\theta\right) \mathmr{and} x = r \cos \left(\theta\right)$, and then solve for r.

#### Explanation:

Given: ${\left(x - 1\right)}^{2} - {\left(y + 5\right)}^{2} = - 24$

Here is a graph of the above equation:

Convert to polar coordinates.

Expand the squares:

${x}^{2} - 2 x + 1 - \left({y}^{2} + 10 y + 25\right) = - 24$

Regroup by power:

${x}^{2} - {y}^{2} - 2 x - 10 y + 1 - 25 = - 24$

Combine the constant terms:

${x}^{2} - {y}^{2} - 2 x - 10 y = 0$

Substitute $r \cos \left(\theta\right)$ for x and $r \sin \left(\theta\right)$ for y:

${\left(r \cos \left(\theta\right)\right)}^{2} - {\left(r \sin \left(\theta\right)\right)}^{2} - 2 \left(r \cos \left(\theta\right)\right) - 10 \left(r \sin \left(\theta\right)\right) = 0$

Lets move the factors of r outside the ():

$\left({\cos}^{2} \left(\theta\right) - {\sin}^{2} \left(\theta\right)\right) {r}^{2} - \left(2 \cos \left(\theta\right) + 10 \sin \left(\theta\right)\right) r = 0$

There are two roots, $r = 0$ which is trivial should be discarded, and:

$\left({\cos}^{2} \left(\theta\right) - {\sin}^{2} \left(\theta\right)\right) r - \left(2 \cos \left(\theta\right) + 10 \sin \left(\theta\right)\right) = 0$

Solve for r:

$r = \frac{2 \cos \left(\theta\right) + 10 \sin \left(\theta\right)}{{\cos}^{2} \left(\theta\right) - {\sin}^{2} \left(\theta\right)}$

Here is the graph of the above equation: