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

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

Solve for x

$$x=y$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

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

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

Subtract x^{2} from both sides.

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

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

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Subtract y^{2} from both sides.

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

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

$$-4x=-4y$$

Cancel out -4 on both sides.

$$x=y$$

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

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

Simplify. The value x=y satisfies the equation.

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

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

$$x=y$$

Solve for y

$$y=x$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

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

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

Subtract y^{2} from both sides.

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

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

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract x^{2} from both sides.

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

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

$$-4y+4=-4x+4$$

Subtract 4 from both sides.

$$-4y=-4x+4-4$$

Subtract 4 from 4 to get 0.

$$-4y=-4x$$

Cancel out -4 on both sides.

$$y=x$$

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

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

Simplify. The value y=x satisfies the equation.

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

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

$$y=x$$