Convert to Polar x^2+y^2=9

$x^{2} + y^{2} = 9$

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

${(r \cos (θ))}^{2} + y^{2} = 9$

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

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

Solve for $r$ .

Subtract $9$ from both sides of the equation.

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

Simplify the left side of the equation.

Simplify each term.

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

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

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

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

Simplify with factoring out.

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

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

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

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

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

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

Rearrange terms.

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

Apply pythagorean identity.

$r^{2} \cdot 1 - 9 = 0$

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

$r^{2} - 9 = 0$

Factor $r^{2} - 9$ .

Rewrite $9$ as $3^{2}$ .

$r^{2} - 3^{2} = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = r$ and $b = 3$ .

$(r + 3) (r - 3) = 0$

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

$r + 3 = 0$

$r - 3 = 0$

Set $r + 3$ equal to $0$ and solve for $r$ .

Set $r + 3$ equal to $0$ .

$r + 3 = 0$

Subtract $3$ from both sides of the equation.

$r = - 3$

Set $r - 3$ equal to $0$ and solve for $r$ .

Set $r - 3$ equal to $0$ .

$r - 3 = 0$

Add $3$ to both sides of the equation.

$r = 3$

The final solution is all the values that make $(r + 3) (r - 3) = 0$ true.

$r = - 3, 3$

Convert to Polar x^2+y^2=9

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