Solve (x-2)^2+(y-2)^2=1(x-(-2))^2+(y-4)

$$( x - 2 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 1 ( x - ( - 2 ) ) ^ { 2 } + ( y - 4 )$$

Solve for x

$$x=\frac{y^{2}-5y+8}{8}$$

Steps for Solving Linear Equation

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

$$x^{2}-4x+4+\left(y-2\right)^{2}=1\left(x-\left(-2\right)\right)^{2}+y-4$$

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

$$x^{2}-4x+4+y^{2}-4y+4=1\left(x-\left(-2\right)\right)^{2}+y-4$$

Add 4 and 4 to get 8.

$$x^{2}-4x+8+y^{2}-4y=1\left(x-\left(-2\right)\right)^{2}+y-4$$

The opposite of -2 is 2.

$$x^{2}-4x+8+y^{2}-4y=1\left(x+2\right)^{2}+y-4$$

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

$$x^{2}-4x+8+y^{2}-4y=1\left(x^{2}+4x+4\right)+y-4$$

Use the distributive property to multiply 1 by x^{2}+4x+4.

$$x^{2}-4x+8+y^{2}-4y=x^{2}+4x+4+y-4$$

Subtract 4 from 4 to get 0.

$$x^{2}-4x+8+y^{2}-4y=x^{2}+4x+y$$

Subtract x^{2} from both sides.

$$x^{2}-4x+8+y^{2}-4y-x^{2}=4x+y$$

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

$$-4x+8+y^{2}-4y=4x+y$$

Subtract 4x from both sides.

$$-4x+8+y^{2}-4y-4x=y$$

Combine -4x and -4x to get -8x.

$$-8x+8+y^{2}-4y=y$$

Subtract 8 from both sides.

$$-8x+y^{2}-4y=y-8$$

Subtract y^{2} from both sides.

$$-8x-4y=y-8-y^{2}$$

Add 4y to both sides.

$$-8x=y-8-y^{2}+4y$$

Combine y and 4y to get 5y.

$$-8x=5y-8-y^{2}$$

The equation is in standard form.

$$-8x=-y^{2}+5y-8$$

Divide both sides by -8.

$$\frac{-8x}{-8}=\frac{-y^{2}+5y-8}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x=\frac{-y^{2}+5y-8}{-8}$$

Divide 5y-8-y^{2} by -8.

$$x=\frac{y^{2}}{8}-\frac{5y}{8}+1$$

Solve for y (complex solution)

$$y=\frac{\sqrt{32x-7}+5}{2}$$

$$y=\frac{-\sqrt{32x-7}+5}{2}$$

Solve for y

$$y=\frac{\sqrt{32x-7}+5}{2}$$

$$y=\frac{-\sqrt{32x-7}+5}{2}\text{, }x\geq \frac{7}{32}$$