multiply both sides by r: 

\(\displaystyle r^2 = 2 r \cos \theta + r \sin \theta\)

Now we can make the substitutions \(\displaystyle r ^2 = x^2 + y^2\), \(\displaystyle r \cos \theta = x\), and \(\displaystyle r \sin \theta = y\):

\(\displaystyle x^2 + y^2 = 2x + y\)

If we subtract the terms on the right from both sides, we can complete the square twice and make this into a regular circle equation:

\(\displaystyle x^2 - 2x \enspace \enspace + y^2 - y \enspace \enspace = 0\)

To complete the square for x, add 1, since \(\displaystyle (\frac{1}{2}(-2))^2= (-1)^ 2 =1\).

To complete the square for y, add \(\displaystyle \frac{1}{4}\) since \(\displaystyle (\frac{1}{2}(-1))^2 = (\frac{1}{2})^2 = \frac{1}{4}\).

\(\displaystyle (x^2 - 2x + 1 ) + (y^2 - y + \frac{1}{4} ) =+ 1+\frac{1}{4}\)

re-write the left side as binomials squared:

\(\displaystyle (x-1)^2 + (y-\frac{1}{2})^2 = 1\frac{1}{4}\)

Multiply both sides by r:

\(\displaystyle r^2 = 8r \sin \theta - 2 r \cos \theta\)

Now we can substitute in \(\displaystyle r^2 = x^2 + y^2\) , \(\displaystyle x= r \cos \theta\), and \(\displaystyle y = r \sin \theta\).

\(\displaystyle x^2 + y^2 = 8y-2x\) if we subtract both terms on the right from both sides, we can complete the square twice to get this into the normal form for a circle.

\(\displaystyle x^2 + 2x \enspace \enspace + y^2 -8y \enspace \enspace =0\)

To complete the square for x, add 1 to both sides, since \(\displaystyle (\frac{1}{2}(2))^2 = (1)^2 = 1\)

To complete the square for y, add 16 to both sides, since \(\displaystyle (\frac{1}{2}(-8))^2 = (-4)^2 = 16\):

\(\displaystyle (x^2 + 2x + 1 ) + (y^2 - 8y + 16 ) = +1 + 16\) Re-write the left side as two binomials squared

\(\displaystyle (x+1)^2 + (y-4)^2 = 17\)

To convert, make the substitutions \(\displaystyle r = \sqrt{x^2 + y^2 }\) and \(\displaystyle \tan \theta = \frac{y}{x}\):

\(\displaystyle \sqrt{x^2 +y^2} = \frac{y}{x}\) square both sides

\(\displaystyle x^2 + y^2 = \frac{y^2}{x^2}\) multiply both sides by \(\displaystyle x^2\)

\(\displaystyle x^2(x^2 + y^2 ) = y^2\)

\(\displaystyle x^4 + x^2 y^2 = y^2\) subtract \(\displaystyle y^2\) from both sides

\(\displaystyle x^4 + x^2 y^2 - y^2 = 0\) subtract \(\displaystyle x^4\) from both sides

\(\displaystyle x^2 y^2 - y^2 = -x^4\) factor our \(\displaystyle y^2\)

\(\displaystyle y^2 (x^2 - 1 ) = -x^4\) divide both sides  by \(\displaystyle x^2 - 1\)

\(\displaystyle y ^2 = \frac{ -x^4}{ x^2 - 1 }\) take the square root of both sides

\(\displaystyle y = \sqrt{\frac{-x^4 }{ x^2 - 1 }}\)

Multiply both sides by \(\displaystyle 1 + 7 \sin ^2 \theta\):

\(\displaystyle r(1 + 7 \sin ^2 \theta ) = 8 \cos \theta\) distribute

\(\displaystyle r + 7 r \sin ^2 \theta = 8 \cos \theta\) multiply both sides by r

\(\displaystyle r^2 + 7 r^2 \sin^2 \theta = 8 r \cos \theta\)

convert to rectangular by making the substitutions \(\displaystyle x^2 + y^2 = r^2\) , \(\displaystyle y = r \sin \theta\), and \(\displaystyle x = r \cos \theta\):

\(\displaystyle x^2 + y^2 + 7 y^2 = 8 x\) subtract 8x from both sides and combine like terms

\(\displaystyle x^2 - 8x + 8y^2 = 0\) complete the square for x by adding 16 to both sides

\(\displaystyle x^2 - 8x + 16 + 8y^2 = 16\)

\(\displaystyle (x-4)^2 + 8y^2 = 16\) divide both sides by 16

\(\displaystyle \frac{(x-4)^2}{16 } + \frac{y^2}{2} = 1\)

Write the Cartesian to polar conversion formulas.

\(\displaystyle r^2=x^2+y^2\)

\(\displaystyle \theta=tan^-^1 \left(\frac{y}{x}\right)\)

Substitute the coordinate point to the equations and solve for \(\displaystyle (r,\theta)\).

\(\displaystyle r^2=x^2+y^2\)

\(\displaystyle r=\sqrt{0^2+2^2}=2\)

\(\displaystyle \theta=tan^-^1 \left(\frac{y}{x}\right)=tan^-^1 \left(\frac{2}{0}\right)=\frac{\pi}{2}\)

Since \(\displaystyle (0,2)\) is located in between the first and second quadrant, this is the correct angle.

Therefore, the answer is \(\displaystyle \left(2,\frac{\pi}{2}\right)\).

Write the Cartesian to polar conversion formulas.

\(\displaystyle r^2=x^2+y^2\)

\(\displaystyle \theta=tan^-^1 \left(\frac{y}{x}\right)\)

Substitute the coordinate point to the equations to find \(\displaystyle (r,\theta)\).

\(\displaystyle r^2=x^2+y^2\)

\(\displaystyle r=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt2\)

\(\displaystyle \theta=tan^-^1 \left(\frac{y}{x}\right)=tan^-^1 \left(\frac{-3}{-3}\right)=\frac{\pi}{4}\)

Since \(\displaystyle (-3,-3)\) is not located in between the first quadrant, this is not the correct angle.  The correct location of this coordinate is in the third quadrant. Add \(\displaystyle \pi\) radians to get the correct angle.

\(\displaystyle \theta=\frac{\pi}{4}+\pi=\frac{5\pi}{4}\)

Therefore, the answer is \(\displaystyle \left(3\sqrt2,\frac{5\pi}{4}\right)\).

Because \(\displaystyle x=r\cos\theta\) and \(\displaystyle y=r\sin\theta\), substitute those values in the rectangular form.

\(\displaystyle (x+4)^2+y^2=16\)

\(\displaystyle (r\cos\theta+4)^2+(r\sin\theta)^2=16\)

Now, expand the equation.

\(\displaystyle r^2\cos^2\theta+8r\cos\theta+16+r^2\sin^2\theta=16\)

Subtract \(\displaystyle 16\) from both sides.

\(\displaystyle r^2\cos^2\theta+8r\cos\theta+r^2\sin^2\theta=0\)

\(\displaystyle r^2(\cos^2\theta+\sin^2\theta)+8r\cos\theta=0\)

Recall the trigonometric identity \(\displaystyle \sin^2\theta+\cos^2\theta=1\)

\(\displaystyle r^2+8r\cos\theta=0\)

Factor the equation.

\(\displaystyle r(r+8\cos\theta)=0\)

Recall that \(\displaystyle x=r\cos\theta\) and \(\displaystyle y=r\sin\theta\).

Substitute these values into the equation.

\(\displaystyle r\sin\theta=3(r\cos\theta)+2\)

Now, manipulate the equation so that the terms with \(\displaystyle r\) are on the same side.

\(\displaystyle r\sin\theta-3r\cos\theta=2\)

Factor out the \(\displaystyle r\).

\(\displaystyle r(\sin\theta-3\cos\theta)=2\)

Divide both sides by \(\displaystyle \sin\theta-3\cos\theta\).

\(\displaystyle r=\frac{2}{\sin\theta-3\cos\theta}\)

Recall that \(\displaystyle x=r\cos\theta\) and \(\displaystyle y=r\sin\theta\).

Substitute these values into the equation.

\(\displaystyle r\sin\theta=-2(r\cos\theta)-5\)

Now, manipulate the equation so that the terms with \(\displaystyle r\) are on the same side.

\(\displaystyle r\sin\theta+2r\cos\theta=-5\)

Factor out the \(\displaystyle r\).

\(\displaystyle r(\sin\theta+2\cos\theta)=-5\)

Divide both sides by \(\displaystyle \sin\theta+2\cos\theta\).

\(\displaystyle r=-\frac{5}{\sin\theta+2\cos\theta}\)

Recall that \(\displaystyle x=r\cos\theta\) and \(\displaystyle y=r\sin\theta\).

Substitute these values into the equation.

\(\displaystyle r\cos\theta=7(r\sin\theta)+2\)

Now, manipulate the equation so that the terms with \(\displaystyle r\) are on the same side.

\(\displaystyle r\cos\theta-7r\sin\theta=2\)

Factor out the \(\displaystyle r\).

\(\displaystyle r(\cos\theta-7\sin\theta)=2\)

Divide both sides by \(\displaystyle \cos\theta-7\sin\theta\).

\(\displaystyle r=\frac{2}{\cos\theta-7\sin\theta}\)