Question 172172
 i need to convert this rectangular equation into a polar equation: the circle centered at (0,-4), with a radius of 4.
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The standard equation for a circle in
rectangular coordinates with center (h,k)
and radius r is:

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

Substitute h=0, k=-4, and r=4

{{{(x-0)^2+(y-(-4))^2=(4)^2}}}

{{{x^2+(y+4)^2=16}}}

{{{x^2+y^2+8y+16=16}}}

{{{x^2+y^2+8y=0}}}

Now we draw a typical angle {{{phi}}} in standard 
position. Unfortunately the software here can't
get the Greek letter theta, so I'll use "phi" instead. 

{{{drawing(400,400,-5,5,-5,5,
graph(400,400,-5,5,-5,5),locate(2.8,1.8,y),
triangle(0,0,2.7,3.4,2.7,0), locate(1.5,.5,x),
locate(1.1,2,r), locate(.4,.6,phi)  )}}}

Write down the trig equation for the sine,
cosine, and and the Pythagorean theorem:

{{{matrix(2,3,

sin(phi)=y/r, "",cos(phi)=x/r, 

"", x^2+y^2=r^2, "")}}}

Solve the first two equations for 
y and x:

{{{matrix(2,3,

y= r*sin(phi), "",x=r*cos(phi), 

"", x^2+y^2=r^2, "")}}}

Now go back to this equation:

{{{x^2+y^2+8y=0}}}

Notice we can replace the first two
terms {{{x^2+y^2}}} by {{{r^2}}}
and we get:

{{{r^2+8y=0}}}

Now we can replace {{{y}}} by
{{{r*sin(phi)}}}

{{{r^2+8y=0}}}

{{{r^2+8r*sin(phi)=0}}}

Factor out r:

{{{r(r+8sin(phi))=0}}}

{{{r=0}}}, {{{r+8sin(phi)=0}}}

The first equation can be ignored. It is
just the equation of the origin (pole).
But the circle contains the origin (pole).
So we end up with:

{{{r+8sin(phi)=0}}}

Edwin</pre>