# How do you graph the equations (x - 2)^2 + (y + 2)^2 = 8 and r = 4 \cos \theta - 4 \sin \theta?

This two equations give birth to the same graph: a circumference centered in $C = \left(2 , - 2\right)$ with radius $r = 2 \sqrt{2}$.

The equation ${\left(x - 2\right)}^{2} + {\left(y + 2\right)}^{2} = 8$ is the result of a scaling by a factor of $2 \sqrt{2}$ and the a translation of vector $\vec{v} = \left(2 , - 2\right)$, applied in this order to a unit circle ${x}^{2} + {y}^{2} = 1$.
In fact, the scaling equations are
$\left(x \text{'",y"'}\right) = 2 \sqrt{2} \left(x , y\right) = \left(2 \sqrt{2} x , 2 \sqrt{2} y\right)$
and the translation equations are
$\left(x \text{''",y"''")=(x"'"+2,y"'} - 2\right)$

To find out that the result of this transformation is the equation ${\left(x \text{''"-2)^2+(y"''} + 2\right)}^{2} = 8$, we have to invert these transformations. So we have the inverse translation $\left(x \text{'",y"'")=(x"''"-2,y"''} + 2\right)$ and the inverse scaling $\left(x , y\right) = \frac{1}{2 \sqrt{2}} \left(x \text{'",y"'}\right)$.
Now we can compose them, getting (x,y) = 1/{2 sqrt{2}} (x"'",y"'") = 1/{2 sqrt{2}} (x"''"-2,y"''"+2)=({x"''"-2}/{2 sqrt{2}},{y"''"+2}/{2 sqrt{2}}).
We can substitute these into the unit circle equation, getting
${\left(\frac{x \text{''"-2}/{2 sqrt{2}})^2+({y"''} + 2}{2 \sqrt{2}}\right)}^{2} = 1$
${\left(x \text{''"-2)^2/8+(y"''} + 2\right)}^{2} / 8 = 1$
${\left(x \text{''"-2)^2+(y"''} + 2\right)}^{2} = 8$

This shows that the first equation is a circle of center $C = \left(2 , - 2\right)$ and radius $r = 2 \sqrt{2}$.

To show that the polar equation $r = 4 \cos \theta - 4 \sin \theta$ gives the same graph, we simply convert polar coordinates to cartesian coordinates ($r = \sqrt{{x}^{2} + {y}^{2}}$ and $\theta = \arctan \left(\frac{y}{x}\right)$) to show that this is the polar form of the equation ${\left(x - 2\right)}^{2} + {\left(y + 2\right)}^{2} = 8$. If we substitute $r$ and $\theta$ in the polar equation, we get:
$\sqrt{{x}^{2} + {y}^{2}} = 4 \cos \left(\arctan \left(\frac{y}{x}\right)\right) - 4 \sin \left(\arctan \left(\frac{y}{x}\right)\right)$.

Now consider a right triangle with the two legs $\overline{A C} = x$ and $\overline{B C} = y$, so $\overline{A B} = \sqrt{{x}^{2} + {y}^{2}}$.

$\sin \theta = \frac{\overline{B C}}{\overline{A B}} = \frac{y}{\sqrt{{x}^{2} + {y}^{2}}}$
$\cos \theta = \frac{\overline{A C}}{\overline{A B}} = \frac{x}{\sqrt{{x}^{2} + {y}^{2}}}$
$\tan \theta = \frac{y}{x} \Rightarrow \theta = \arctan \left(\frac{y}{x}\right)$
So we get
$\frac{y}{\sqrt{{x}^{2} + {y}^{2}}} = \sin \theta = \sin \left(\arctan \left(\frac{y}{x}\right)\right)$
$\frac{x}{\sqrt{{x}^{2} + {y}^{2}}} = \cos \theta = \cos \left(\arctan \left(\frac{y}{x}\right)\right)$
And these equalities are valid for non-negative values of $x$ and $y$ because of the symmetry properties of the functions sine, cosine and arctangent.

So we can rewrite the cartesian equation that we got from the polar one, getting:
$\sqrt{{x}^{2} + {y}^{2}} = 4 \frac{x}{\sqrt{{x}^{2} + {y}^{2}}} - 4 \frac{y}{\sqrt{{x}^{2} + {y}^{2}}}$
${x}^{2} + {y}^{2} = 4 x - 4 y$
${x}^{2} - 4 x + {y}^{2} + 4 y = 0$
Finally, we complete the squares:
${x}^{2} - 4 x + 4 + {y}^{2} + 4 y + 4 = 4 + 4$
${\left(x - 2\right)}^{2} + {\left(y + 2\right)}^{2} = 8$