# How do you convert 1=(-x+4)^2+(2y+9)^2 into polar form?

${r}^{2} \left({\cos}^{2} \theta + 4 {\sin}^{2} \theta\right) - r \left(4 \cos \theta - 18 \sin \theta\right) + 40 = 0$

#### Explanation:

Given:
$1 = {\left(- x + 4\right)}^{2} + {\left(2 y + 9\right)}^{2}$

$1 = {\left(- x + 4\right)}^{2} + {\left(2 \left(y + \frac{9}{2}\right)\right)}^{2}$
$1 = {\left(- x + 4\right)}^{2} + 4 {\left(y + \frac{9}{2}\right)}^{2}$

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

Substituting
$1 = {\left(- r \cos \theta + 4\right)}^{2} + {\left(2 r \sin \theta + 9\right)}^{2}$

$1 = {r}^{2} {\cos}^{2} \theta - 8 r \cos \theta + 16 + 4 {r}^{2} {\sin}^{2} \theta + 36 r \sin \theta + 81$

${r}^{2} \left({\cos}^{2} \theta + 4 {\sin}^{2} \theta\right) + r \left(- 8 \cos \theta + 36 \sin \theta\right) + 81 - 1 = 0$

$2 {r}^{2} \left({\cos}^{2} \theta + 4 {\sin}^{2} \theta\right) - 2 r \left(4 \cos \theta - 18 \sin \theta\right) + 2 \left(40\right) = 0$

${r}^{2} \left({\cos}^{2} \theta + 4 {\sin}^{2} \theta\right) - r \left(4 \cos \theta - 18 \sin \theta\right) + 40 = 0$