Convert to Polar x^2+(y-3)^2=9

$x^{2} + {(y - 3)}^{2} = 9$

Step 1

Since $x = r \cos (θ)$ , replace $x$ with $r \cos (θ)$ .

${(r \cos (θ))}^{2} + {(y - 3)}^{2} = 9$

Step 2

Since $y = r \sin (θ)$ , replace $y$ with $r \sin (θ)$ .

${(r \cos (θ))}^{2} + {((r \sin (θ)) - 3)}^{2} = 9$

Step 3

Solve for $r$ .

Subtract $9$ from both sides of the equation.

${(r \cos (θ))}^{2} + {((r \sin (θ)) - 3)}^{2} - 9 = 0$

Simplify the left side of the equation.

Simplify each term.

Apply the product rule to $r \cos (θ)$ .

$r^{2} \cos^{2} (θ) + {((r \sin (θ)) - 3)}^{2} - 9 = 0$

Rewrite ${(r \sin (θ) - 3)}^{2}$ as $(r \sin (θ) - 3) (r \sin (θ) - 3)$ .

$r^{2} \cos^{2} (θ) + (r \sin (θ) - 3) (r \sin (θ) - 3) - 9 = 0$

Expand $(r \sin (θ) - 3) (r \sin (θ) - 3)$ using the FOIL Method.

Apply the distributive property.

$r^{2} \cos^{2} (θ) + r \sin (θ) (r \sin (θ) - 3) - 3 (r \sin (θ) - 3) - 9 = 0$

Apply the distributive property.

$r^{2} \cos^{2} (θ) + r \sin (θ) (r \sin (θ)) + r \sin (θ) \cdot - 3 - 3 (r \sin (θ) - 3) - 9 = 0$

Apply the distributive property.

$r^{2} \cos^{2} (θ) + r \sin (θ) (r \sin (θ)) + r \sin (θ) \cdot - 3 - 3 (r \sin (θ)) - 3 \cdot - 3 - 9 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $r$ by $r$ by adding the exponents.

Move $r$ .

$r^{2} \cos^{2} (θ) + r \cdot r \sin (θ) \sin (θ) + r \sin (θ) \cdot - 3 - 3 (r \sin (θ)) - 3 \cdot - 3 - 9 = 0$

Multiply $r$ by $r$ .

$r^{2} \cos^{2} (θ) + r^{2} \sin (θ) \sin (θ) + r \sin (θ) \cdot - 3 - 3 (r \sin (θ)) - 3 \cdot - 3 - 9 = 0$

Multiply $r^{2} \sin (θ) \sin (θ)$ .

Raise $\sin (θ)$ to the power of $1$ .

$r^{2} \cos^{2} (θ) + r^{2} (\sin (θ) \sin (θ)) + r \sin (θ) \cdot - 3 - 3 (r \sin (θ)) - 3 \cdot - 3 - 9 = 0$

Raise $\sin (θ)$ to the power of $1$ .

$r^{2} \cos^{2} (θ) + r^{2} (\sin (θ) \sin (θ)) + r \sin (θ) \cdot - 3 - 3 (r \sin (θ)) - 3 \cdot - 3 - 9 = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$r^{2} \cos^{2} (θ) + r^{2} {\sin (θ)}^{1 + 1} + r \sin (θ) \cdot - 3 - 3 (r \sin (θ)) - 3 \cdot - 3 - 9 = 0$

Add $1$ and $1$ .

$r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) + r \sin (θ) \cdot - 3 - 3 (r \sin (θ)) - 3 \cdot - 3 - 9 = 0$

Move $- 3$ to the left of $r \sin (θ)$ .

$r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) - 3 \cdot (r \sin (θ)) - 3 (r \sin (θ)) - 3 \cdot - 3 - 9 = 0$

Multiply $- 3$ by $- 3$ .

$r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) - 3 r \sin (θ) - 3 r \sin (θ) + 9 - 9 = 0$

Subtract $3 r \sin (θ)$ from $- 3 r \sin (θ)$ .

$r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) - 6 r \sin (θ) + 9 - 9 = 0$

Simplify terms.

Combine the opposite terms in $r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) - 6 r \sin (θ) + 9 - 9$ .

Subtract $9$ from $9$ .

$r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) - 6 r \sin (θ) + 0 = 0$

Add $r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) - 6 r \sin (θ)$ and $0$ .

$r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) - 6 r \sin (θ) = 0$

Factor $r^{2}$ out of $r^{2} \cos^{2} (θ)$ .

$r^{2} (\cos^{2} (θ)) + r^{2} \sin^{2} (θ) - 6 r \sin (θ) = 0$

Factor $r^{2}$ out of $r^{2} \sin^{2} (θ)$ .

$r^{2} (\cos^{2} (θ)) + r^{2} (\sin^{2} (θ)) - 6 r \sin (θ) = 0$

Factor $r^{2}$ out of $r^{2} (\cos^{2} (θ)) + r^{2} (\sin^{2} (θ))$ .

$r^{2} (\cos^{2} (θ) + \sin^{2} (θ)) - 6 r \sin (θ) = 0$

Rearrange terms.

$r^{2} (\sin^{2} (θ) + \cos^{2} (θ)) - 6 r \sin (θ) = 0$

Apply pythagorean identity.

$r^{2} \cdot 1 - 6 r \sin (θ) = 0$

Multiply $r^{2}$ by $1$ .

$r^{2} - 6 r \sin (θ) = 0$

Factor $r$ out of $r^{2} - 6 r \sin (θ)$ .

Factor $r$ out of $r^{2}$ .

$r \cdot r - 6 r \sin (θ) = 0$

Factor $r$ out of $- 6 r \sin (θ)$ .

$r \cdot r + r (- 6 \sin (θ)) = 0$

Factor $r$ out of $r \cdot r + r (- 6 \sin (θ))$ .

$r (r - 6 \sin (θ)) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$r = 0$

$r - 6 \sin (θ) = 0$

Set $r$ equal to $0$ .

$r = 0$

Set $r - 6 \sin (θ)$ equal to $0$ and solve for $r$ .

Set $r - 6 \sin (θ)$ equal to $0$ .

$r - 6 \sin (θ) = 0$

Add $6 \sin (θ)$ to both sides of the equation.

$r = 6 \sin (θ)$

The final solution is all the values that make $r (r - 6 \sin (θ)) = 0$ true.

$r = 0$

$r = 6 \sin (θ)$

$r = 0$

$r = 6 \sin (θ)$

Convert to Polar x^2+(y-3)^2=9

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