Question 983185
<pre>
I won't do yours for you.  I'll do one exactly like it instead
and you can use it as a model to do yours by.  I'll do this
one instead which uses cosine instead of sine:

{{{r-4cos(theta)}}}{{{""=""}}}{{{0}}}

Always draw this picture when converting from polar to ractangular 
or from rectangular to polar::

{{{drawing(200,120,-1,1,-.6,.6,line(0,2,0,-2),line(2,0,-2,0),
triangle(0,0,.6,0,.6,.4), locate(.3,0,x), locate(.62,.3,y),
locate(.27,.36,r), red(arc(0,0,.5,-.5,0,34)) locate(.2,.17,theta) )}}}

Looking at the triangle in the picture above you can see these:

{{{y/r=sin(theta)}}}, {{{x/r=cos(theta)}}}, {{{r^2=x^2+y^2}}}

Then solve for y, x and r

{{{y=r*sin(theta)}}}, {{{x=r*cos(theta)}}},  {r=sqrt(x^2+y^2)}}}

ALWAYS WAIT TILL LAST TO SUBSTITUTE FOR r. Substitute for x and y
first. Then simplify.

{{{r-4cos(theta)}}}{{{""=""}}}{{{0}}}

{{{r-4(x/r)}}}{{{""=""}}}{{{0}}}

Multiply through by r

{{{r^2-4x}}}{{{""=""}}}{{{0}}}

NOW, FINALLY, we substitute for r<sup>2</sup>.

{{{x^2+y^2-4x}}}{{{""=""}}}{{{0}}}   <-- answer

That's it!.

Now do yours the same way, only use the sine instead.

Edwin</pre>