Question 865281
Find the polar coordinates of the point whose rectangular coordinates are 
({{{4sqrt(3)}}}, -4)
<pre>
We will indicate the rectangular coordinates of point P in 
black as {{{P(x,y)}}} and the polar coordinates of point P
in red as {{{red(P(r,theta))}}}

Plot P using its rectangular coordinates {{{P(x,y)}}} = ({{{P(4sqrt(3),0)}}}:

{{{drawing(400,400,-8,8,-8,8, graph(400,400,-8,8,-8,8),
locate(4.5,-4.4,P(4sqrt(3),-4)),
circle(6.92820323,-4,0.15),circle(6.92820323,-4,0.13),circle(6.92820323,-4,0.11),circle(6.92820323,-4,0.09),circle(6.92820323,-4,0.07),circle(6.92820323,-4,0.05),circle(6.92820323,-4,0.03),circle(6.92820323,-4,0.01))}}}

The x-coordinate of point P is {{{x=4sqrt(3)}}} and the y-coordinate
of point P is -4.

Draw a right triangle with this point P and the origin as
vertices and the right angle on the x-axis.  The legs of this right
triangle x and y are the RECTANGULAR coordinates of point P. The 
hypotenuse r of this right triangle is the first POLAR coordinate 
of P.  The angle {{{theta}}} indicated by the counter-clockwise red 
arc is the second POLAR coordinate of the point {{{red(P(r,theta))}}} 

{{{drawing(400,400,-8,8,-8,8, graph(400,400,-8,8,-8,8),
locate(4.5,-4.4,P(4sqrt(3),-4)),locate(3.8,1,x=4sqrt(3)),
locate(6,-1.7,y=-4),red(locate(3.5,-2.2,r)), red(locate(-1.5,1.5,theta)),

red(arc(0,0,3,-3,0,330)),
circle(6.92820323,-4,0.15),circle(6.92820323,-4,0.13),circle(6.92820323,-4,0.11),circle(6.92820323,-4,0.09),circle(6.92820323,-4,0.07),circle(6.92820323,-4,0.05),circle(6.92820323,-4,0.03),circle(6.92820323,-4,0.01),
green(triangle(0,0,4sqrt(3),0,4sqrt(3),-4))



)}}}

We only need to calculate {{{red(r)}}} and {{{ theta}}}

{{{red(r)^2=x^2+y^2}}}
{{{red(r)^2=(4sqrt(3))^2+(-4)^2}}}
{{{red(r)^2=16*3+16}}}
{{{red(r)^2=48+16}}}
{{{red(r)^2=64}}}
{{{red(r)=8}}}

{{{tan(red(theta))=y/x=(-4)/(4sqrt(3))=-1/sqrt(3)}}}

Therefore {{{red(theta)}}} in the 4th quadrant is {{{red(11pi/6)}}}, and

{{{P(4sqrt(3),0)}}} = {{{red(P(8,11pi/6))}}}

{{{drawing(400,400,-8,8,-8,8, graph(400,400,-8,8,-8,8),
red(locate(4.5,-4.4,P(8,11pi/6))),locate(3.8,1,x=4sqrt(3)),
locate(6,-1.7,y=-4),red(locate(2.5,-2.2,r=8)), red(locate(-3.5,1.5,theta=11pi/6)),

red(arc(0,0,3,-3,0,330)),
circle(6.92820323,-4,0.15),circle(6.92820323,-4,0.13),circle(6.92820323,-4,0.11),circle(6.92820323,-4,0.09),circle(6.92820323,-4,0.07),circle(6.92820323,-4,0.05),circle(6.92820323,-4,0.03),circle(6.92820323,-4,0.01),
green(triangle(0,0,4sqrt(3),0,4sqrt(3),-4))



)}}}
</pre>
Find the rectangular coordinates of the point whose polar coordinates are 
{{{red(P(-4,pi/6))}}}.
<pre>
Let's draw the point {{{red(P(-4,pi/6))}}} using its polar coordinates.

First we draw the angle {{{red(theta)}}} with a dotted line through
the origin:

{{{drawing(400,400,-8,8,-8,8, graph(400,400,-8,8,-8,8,

( sqrt(3)/3 )x*(sqrt(sin(9x))/sqrt(sin(9x))) ),
red(arc(0,0,3,-3,0,30),locate(2,1.6,theta=pi/6))
 )}}}

Next we locate the value of r on the x-axis, then we swing an
arc from that point on the x-axis around to the dotted line,
like the green arc below swinging from -4 on the x-axis to
the dotted line, and mar that point {{{red(P(-4,pi/6))}}}.

{{{drawing(400,400,-8,8,-8,8, graph(400,400,-8,8,-8,8,

( sqrt(3)/3 )x*(sqrt(sin(9x))/sqrt(sin(9x))) ),
red(arc(0,0,3,-3,0,30),locate(2,1.6,theta=pi/6)),
green(arc(0,0,8,-8,180,210)),

circle(-3.46410162,-2,0.15),circle(-3.46410162,-2,0.13),circle(-3.46410162,-2,0.11),circle(-3.46410162,-2,0.09),circle(-3.46410162,-2,0.07),circle(-3.46410162,-2,0.05),circle(-3.46410162,-2,0.03),circle(-3.46410162,-2,0.01),

red(locate(-4,-2,P(-4,pi/6)))
 )}}}

Then we erase the green arc and draw a right triangle with 
this point P and the origin as vertices and the right angle 
on the x-axis, and indicate the RECTANGULAR coordinates x,
y, of the point P.  Since we swung the point from the point
x=-4, the hypotenuse {{{red(r=-4)}}}.

{{{drawing(400,400,-8,8,-8,8, graph(400,400,-8,8,-8,8,

( sqrt(3)/3 )x*(sqrt(sin(9x))/sqrt(sin(9x))) ),
red(arc(0,0,3,-3,0,30),locate(2,1.6,theta=pi/6)),

red(locate(-2,-1,r=-4)),


circle(-3.46410162,-2,0.15),circle(-3.46410162,-2,0.13),circle(-3.46410162,-2,0.11),circle(-3.46410162,-2,0.09),circle(-3.46410162,-2,0.07),circle(-3.46410162,-2,0.05),circle(-3.46410162,-2,0.03),circle(-3.46410162,-2,0.01),
green(triangle(0,0,-3.46410162,0,-3.46410162,-2)),
locate(-2,.8,x), locate(-4,-.6,y),
red(locate(-4,-2,P(-4,pi/6)))
 )}}}  

Now we calculate x and y.

{{{cos(theta)=x/red(r)}}}
{{{cos(pi/6)=x/red(-4)}}}
{{{sqrt(3)/2=x/red(-4)}}}
{{{2x=red(-4)sqrt(3)}}}
{{{x=-2sqrt(3)}}}

{{{sin(theta)=y/red(r)}}}
{{{sin(pi/6)=y/red(-4)}}}
{{{y=red(-4)sin(pi/6)}}}
{{{y=red(-4)(1/2)}}}
{{{y=-2}}}

So {{{red(P(-4,pi/6))}}} = {{{P(-2sqrt(3),-2)}}}

{{{drawing(400,400,-8,8,-8,8, graph(400,400,-8,8,-8,8,

( sqrt(3)/3 )x*(sqrt(sin(9x))/sqrt(sin(9x))) ),
red(arc(0,0,3,-3,0,30),locate(2,1.6,theta=pi/6)),

red(locate(-2,-1,r=-4)),


circle(-3.46410162,-2,0.15),circle(-3.46410162,-2,0.13),circle(-3.46410162,-2,0.11),circle(-3.46410162,-2,0.09),circle(-3.46410162,-2,0.07),circle(-3.46410162,-2,0.05),circle(-3.46410162,-2,0.03),circle(-3.46410162,-2,0.01),
green(triangle(0,0,-3.46410162,0,-3.46410162,-2)),
locate(-3.2,1.1,x=-2sqrt(3)), locate(-5.3,-.8,y=-2),
locate(-4,-2.3,P(-2sqrt(3),-2))
 )}}}
</pre>
Find a rectangular form of the equation {{{red(r = 5 cos(theta))}}}
<pre>
Always substitute trig functions first, 

{{{sin(red(theta))=y/red(r)}}}, {{{cos(red(theta))=x/red(r)}}}, {{{tan(red(theta))=y/x}}}

and always wait until last to replace {{{red(r)}}} by {{{red(r)=sqrt(x^2+y^2)}}}

{{{red(r)= 5cos(theta)}}}
{{{red(r) = 5(x/red(r))}}}
{{{red(r) = 5x/red(r)}}}
{{{red(r^2) = 5x}}}
{{{(sqrt(x^2+y^2))^2 = 5x}}}
{{{x^2+y^2=5x}}}

You can then put it in standard form of a circle,

{{{(x-h)^2+(y-k)^2=r^2}}}

getting:

{{{(x-5/2)^2+(y-0)^2=(5/2)^2}}}

whose graph is a circle with center (h,k) = ({{{5/2}}},0) 

and radius {{{r=5/2}}}: 

{{{drawing(400,2400/7,-1,6,-3,3, graph(400,2400/7,-1,6,-3,3),
circle(2.5,0,2.5),
circle(2.5,0,0.09),
circle(2.5,0,0.07),circle(2.5,0,0.05),circle(2.5,0,0.03),circle(2.5,0,0.01))}}}

Edwin</pre>