Solve sqrt{(x-2)^2+(y-2)^2}=sqrt{(x-(-2))^2+(y-4)^2}

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

Solve for x

$$x=\frac{y-3}{2}$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Add 4 and 4 to get 8.

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

Calculate \sqrt{x^{2}-4x+8+y^{2}-4y} to the power of 2 and get x^{2}-4x+8+y^{2}-4y.

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

The opposite of -2 is 2.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-4\right)^{2}.

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{x^{2}+4x+4+y^{2}-8y+16}\right)^{2}$$

Add 4 and 16 to get 20.

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{x^{2}+4x+20+y^{2}-8y}\right)^{2}$$

Calculate \sqrt{x^{2}+4x+20+y^{2}-8y} to the power of 2 and get x^{2}+4x+20+y^{2}-8y.

$$x^{2}-4x+8+y^{2}-4y=x^{2}+4x+20+y^{2}-8y$$

Subtract x^{2} from both sides.

$$x^{2}-4x+8+y^{2}-4y-x^{2}=4x+20+y^{2}-8y$$

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

$$-4x+8+y^{2}-4y=4x+20+y^{2}-8y$$

Subtract 4x from both sides.

$$-4x+8+y^{2}-4y-4x=20+y^{2}-8y$$

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

$$-8x+8+y^{2}-4y=20+y^{2}-8y$$

Subtract 8 from both sides.

$$-8x+y^{2}-4y=20+y^{2}-8y-8$$

Subtract 8 from 20 to get 12.

$$-8x+y^{2}-4y=12+y^{2}-8y$$

Subtract y^{2} from both sides.

$$-8x-4y=12+y^{2}-8y-y^{2}$$

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

$$-8x-4y=12-8y$$

Add 4y to both sides.

$$-8x=12-8y+4y$$

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

$$-8x=12-4y$$

Divide both sides by -8.

$$\frac{-8x}{-8}=\frac{12-4y}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x=\frac{12-4y}{-8}$$

Divide 12-4y by -8.

$$x=\frac{y-3}{2}$$

Substitute \frac{y-3}{2} for x in the equation \sqrt{\left(x-2\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}.

$$\sqrt{\left(\frac{y-3}{2}-2\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(\frac{y-3}{2}-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}$$

Simplify. The value x=\frac{y-3}{2} satisfies the equation.

$$\frac{1}{2}\left(65-30y+5y^{2}\right)^{\frac{1}{2}}=\frac{1}{2}\left(65-30y+5y^{2}\right)^{\frac{1}{2}}$$

Equation \sqrt{\left(x-2\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(y-4\right)^{2}+\left(x-\left(-2\right)\right)^{2}} has a unique solution.

$$x=\frac{y-3}{2}$$

Solve for y

$$y=2x+3$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Add 4 and 4 to get 8.

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

Calculate \sqrt{x^{2}-4x+8+y^{2}-4y} to the power of 2 and get x^{2}-4x+8+y^{2}-4y.

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

The opposite of -2 is 2.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-4\right)^{2}.

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{x^{2}+4x+4+y^{2}-8y+16}\right)^{2}$$

Add 4 and 16 to get 20.

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{x^{2}+4x+20+y^{2}-8y}\right)^{2}$$

Calculate \sqrt{x^{2}+4x+20+y^{2}-8y} to the power of 2 and get x^{2}+4x+20+y^{2}-8y.

$$x^{2}-4x+8+y^{2}-4y=x^{2}+4x+20+y^{2}-8y$$

Subtract y^{2} from both sides.

$$x^{2}-4x+8+y^{2}-4y-y^{2}=x^{2}+4x+20-8y$$

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

$$x^{2}-4x+8-4y=x^{2}+4x+20-8y$$

Add 8y to both sides.

$$x^{2}-4x+8-4y+8y=x^{2}+4x+20$$

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

$$x^{2}-4x+8+4y=x^{2}+4x+20$$

Subtract x^{2} from both sides.

$$-4x+8+4y=x^{2}+4x+20-x^{2}$$

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

$$-4x+8+4y=4x+20$$

Add 4x to both sides.

$$8+4y=4x+20+4x$$

Combine 4x and 4x to get 8x.

$$8+4y=8x+20$$

Subtract 8 from both sides.

$$4y=8x+20-8$$

Subtract 8 from 20 to get 12.

$$4y=8x+12$$

Divide both sides by 4.

$$\frac{4y}{4}=\frac{8x+12}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y=\frac{8x+12}{4}$$

Divide 8x+12 by 4.

$$y=2x+3$$

Substitute 2x+3 for y in the equation \sqrt{\left(x-2\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}.

$$\sqrt{\left(x-2\right)^{2}+\left(2x+3-2\right)^{2}}=\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(2x+3-4\right)^{2}}$$

Simplify. The value y=2x+3 satisfies the equation.

$$\left(5+5x^{2}\right)^{\frac{1}{2}}=\left(5+5x^{2}\right)^{\frac{1}{2}}$$

Equation \sqrt{\left(x-2\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(y-4\right)^{2}+\left(x-\left(-2\right)\right)^{2}} has a unique solution.

$$y=2x+3$$