Solve (x-3)^2+(y+5)^2=y^2

$$( x - 3 ) ^ { 2 } + ( y + 5 ) ^ { 2 } = y ^ { 2 }$$

Solve for y

$$y=-\frac{\left(x-3\right)^{2}}{10}-\frac{5}{2}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9+\left(y+5\right)^{2}=y^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+5\right)^{2}.

$$x^{2}-6x+9+y^{2}+10y+25=y^{2}$$

Add 9 and 25 to get 34.

$$x^{2}-6x+34+y^{2}+10y=y^{2}$$

Subtract y^{2} from both sides.

$$x^{2}-6x+34+y^{2}+10y-y^{2}=0$$

Combine y^{2} and -y^{2} to get 0.

$$x^{2}-6x+34+10y=0$$

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

$$-6x+34+10y=-x^{2}$$

Add 6x to both sides.

$$34+10y=-x^{2}+6x$$

Subtract 34 from both sides.

$$10y=-x^{2}+6x-34$$

Divide both sides by 10.

$$\frac{10y}{10}=\frac{-x^{2}+6x-34}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$y=\frac{-x^{2}+6x-34}{10}$$

Divide -x^{2}+6x-34 by 10.

$$y=-\frac{x^{2}}{10}+\frac{3x}{5}-\frac{17}{5}$$

Solve for x (complex solution)

$$x=-\sqrt{-10y-25}+3$$

$$x=\sqrt{-10y-25}+3$$

Solve for x

$$x=-\sqrt{-10y-25}+3$$

$$x=\sqrt{-10y-25}+3\text{, }y\leq -\frac{5}{2}$$