# How do you convert the following to a rectangular equation: r = cos (theta) - sin (theta)?

Oct 5, 2016

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

#### Explanation:

When converting from polar to rectangular equations you have to switch the references to $r$ and $\theta$ to $x$ and $y$.

$x = r \cos \left(\theta\right)$
$y = r \sin \left(\theta\right)$
${r}^{2} = {x}^{2} + {y}^{2}$

Multiply each term by $r$

$r \cdot r = r \cos \left(\theta\right) - r \sin \left(\theta\right)$

Simply

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

Make the necessary substitutions

${x}^{2} + {y}^{2} = x - y$

Gather the $x$ and $y$ variables to the same side

$\left({x}^{2} - x\right) + \left({y}^{2} + y\right) = 0$

Now complete the square. Take the coefficient of both the $x$ and $y$ variable then divide by 2 and then square

${\left(- \frac{1}{2}\right)}^{2} = \frac{1}{4}$

${\left(\frac{1}{2}\right)}^{2} = \frac{1}{4}$

Insert the above 2 quantities on both sides of the equation to keep it balanced.

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

Simplify

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

Convert the 2 perfect square trinomials

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

Check out this tutorial on the conversion of equations from polar to rectangular.

Check out this tutorial on completing the square graphically

Check out this tutorial on completing the square analytically.

Oct 5, 2016

${\left(x - \frac{1}{2}\right)}^{2} + {\left(y + \frac{1}{2}\right)}^{2} = \frac{1}{2}$,
representing the circle through the origin,
with center at $\left(\frac{1}{2} , - \frac{1}{2}\right) \mathmr{and} r a \mathrm{di} u s \frac{1}{\sqrt{2}}$.

#### Explanation:

Use the conversion equation $r \left(\cos \theta , \sin \theta\right) = \left(x , y\right)$ that

gives $r = \sqrt{{x}^{2} + {y}^{2}} , \cos \theta = \frac{x}{r} \mathmr{and} \sin \theta = \frac{y}{r}$.

Here, $r = \sqrt{{x}^{2} + {y}^{2}} = \frac{x - y}{r} = \frac{x - y}{\sqrt{{x}^{2} + {y}^{2}}}$.

Cross multiplying, ${x}^{2} + {y}^{2} = x - y$, or in the standard form,

${\left(x - \frac{1}{2}\right)}^{2} + {\left(y + \frac{1}{2}\right)}^{2} = \frac{1}{2}$, representing the circle through the origin,

with center at $\left(\frac{1}{2} , - \frac{1}{2}\right) \mathmr{and} r a \mathrm{di} u s \frac{1}{\sqrt{2}}$.

Noe that the general polar equation of the circles through the pole r

= 0 is $r = d \cos \left(\theta + \alpha\right)$. Here, the diameter d = sqrt 2 and

$\alpha = \frac{\pi}{4}$..