Find the Expanded Form x^2+(y-3)^2=9

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

Simplify each term.

Rewrite ${(y - 3)}^{2}$ as $(y - 3) (y - 3)$ .

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

Expand $(y - 3) (y - 3)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

$x^{2} + y \cdot y + y \cdot - 3 - 3 y - 3 \cdot - 3 = 9$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

$x^{2} + y^{2} + y \cdot - 3 - 3 y - 3 \cdot - 3 = 9$

Move $- 3$ to the left of $y$ .

$x^{2} + y^{2} - 3 \cdot y - 3 y - 3 \cdot - 3 = 9$

Multiply $- 3$ by $- 3$ .

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

Subtract $3 y$ from $- 3 y$ .

$x^{2} + y^{2} - 6 y + 9 = 9$

Subtract $9$ from both sides of the equation.

$x^{2} + y^{2} - 6 y + 9 - 9 = 0$

Combine the opposite terms in $x^{2} + y^{2} - 6 y + 9 - 9$ .

Subtract $9$ from $9$ .

$x^{2} + y^{2} - 6 y + 0 = 0$

Add $x^{2} + y^{2} - 6 y$ and $0$ .

$x^{2} + y^{2} - 6 y = 0$