This two equations give birth to the same graph: a circumference centered in C=(2,−2) with radius r=2√2.
The equation (x−2)2+(y+2)2=8 is the result of a scaling by a factor of 2√2 and the a translation of vector →v=(2,−2), applied in this order to a unit circle x2+y2=1.
In fact, the scaling equations are
(x',y')=2√2(x,y)=(2√2x,2√2y)
and the translation equations are
(x'',y'')=(x'+2,y'−2)
To find out that the result of this transformation is the equation (x''−2)2+(y''+2)2=8, we have to invert these transformations. So we have the inverse translation (x',y')=(x''−2,y''+2) and the inverse scaling (x,y)=12√2(x',y').
Now we can compose them, getting (x,y)=12√2(x',y')=12√2(x''−2,y''+2)=(x''−22√2,y''+22√2).
We can substitute these into the unit circle equation, getting
(x''−22√2)2+(y''+22√2)2=1
(x''−2)28+(y''+2)28=1
(x''−2)2+(y''+2)2=8
This shows that the first equation is a circle of center C=(2,−2) and radius r=2√2.
To show that the polar equation r=4cosθ−4sinθ gives the same graph, we simply convert polar coordinates to cartesian coordinates (r=√x2+y2 and θ=arctan(yx)) to show that this is the polar form of the equation (x−2)2+(y+2)2=8. If we substitute r and θ in the polar equation, we get:
√x2+y2=4cos(arctan(yx))−4sin(arctan(yx)).
Now consider a right triangle with the two legs ¯¯¯¯¯¯AC=x and ¯¯¯¯¯¯BC=y, so ¯¯¯¯¯¯AB=√x2+y2.
http://www.skwirk.com/p-c_s-12_u-95_t-229_c-764/sine/nsw/sine/trigonometry/elements
sinθ=¯¯¯¯¯¯BC¯¯¯¯¯¯AB=y√x2+y2
cosθ=¯¯¯¯¯¯AC¯¯¯¯¯¯AB=x√x2+y2
tanθ=yx⇒θ=arctan(yx)
So we get
y√x2+y2=sinθ=sin(arctan(yx))
x√x2+y2=cosθ=cos(arctan(yx))
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:
√x2+y2=4x√x2+y2−4y√x2+y2
x2+y2=4x−4y
x2−4x+y2+4y=0
Finally, we complete the squares:
x2−4x+4+y2+4y+4=4+4
(x−2)2+(y+2)2=8