How do you graph the equations (x2)2+(y+2)2=8 and r=4cosθ4sinθ?

1 Answer

This two equations give birth to the same graph: a circumference centered in C=(2,2) with radius r=22.

The equation (x2)2+(y+2)2=8 is the result of a scaling by a factor of 22 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')=22(x,y)=(22x,22y)
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)=122(x',y').
Now we can compose them, getting (x,y)=122(x',y')=122(x''2,y''+2)=(x''222,y''+222).
We can substitute these into the unit circle equation, getting
(x''222)2+(y''+222)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=22.

To show that the polar equation r=4cosθ4sinθ 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 (x2)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/elementshttp://www.skwirk.com/p-c_s-12_u-95_t-229_c-764/sine/nsw/sine/trigonometry/elements
sinθ=¯¯¯¯¯¯BC¯¯¯¯¯¯AB=yx2+y2
cosθ=¯¯¯¯¯¯AC¯¯¯¯¯¯AB=xx2+y2
tanθ=yxθ=arctan(yx)
So we get
yx2+y2=sinθ=sin(arctan(yx))
xx2+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=4xx2+y24yx2+y2
x2+y2=4x4y
x24x+y2+4y=0
Finally, we complete the squares:
x24x+4+y2+4y+4=4+4
(x2)2+(y+2)2=8