Solve sqrt{3+(1-a)^2+(2-b)^2}=sqrt{4+(-2-a)^2+(-2-b)^2}

$$\sqrt { 3 + ( 1 - a ) ^ { 2 } + ( 2 - b ) ^ { 2 } } = \sqrt { 4 + ( - 2 - a ) ^ { 2 } + ( - 2 - b ) ^ { 2 } }$$

Solve for a

$$a=\frac{-4b-2}{3}$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3+\left(1-a\right)^{2}+\left(2-b\right)^{2}}\right)^{2}=\left(\sqrt{4+\left(-2-a\right)^{2}+\left(-2-b\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-a\right)^{2}.

$$\left(\sqrt{3+1-2a+a^{2}+\left(2-b\right)^{2}}\right)^{2}=\left(\sqrt{4+\left(-2-a\right)^{2}+\left(-2-b\right)^{2}}\right)^{2}$$

Add 3 and 1 to get 4.

$$\left(\sqrt{4-2a+a^{2}+\left(2-b\right)^{2}}\right)^{2}=\left(\sqrt{4+\left(-2-a\right)^{2}+\left(-2-b\right)^{2}}\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(2-b\right)^{2}.

$$\left(\sqrt{4-2a+a^{2}+4-4b+b^{2}}\right)^{2}=\left(\sqrt{4+\left(-2-a\right)^{2}+\left(-2-b\right)^{2}}\right)^{2}$$

Add 4 and 4 to get 8.

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

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

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

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

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

Add 4 and 4 to get 8.

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

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(-2-b\right)^{2}.

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

Add 8 and 4 to get 12.

$$8-2a+a^{2}-4b+b^{2}=\left(\sqrt{12+4a+a^{2}+4b+b^{2}}\right)^{2}$$

Calculate \sqrt{12+4a+a^{2}+4b+b^{2}} to the power of 2 and get 12+4a+a^{2}+4b+b^{2}.

$$8-2a+a^{2}-4b+b^{2}=12+4a+a^{2}+4b+b^{2}$$

Subtract 4a from both sides.

$$8-2a+a^{2}-4b+b^{2}-4a=12+a^{2}+4b+b^{2}$$

Combine -2a and -4a to get -6a.

$$8-6a+a^{2}-4b+b^{2}=12+a^{2}+4b+b^{2}$$

Subtract a^{2} from both sides.

$$8-6a+a^{2}-4b+b^{2}-a^{2}=12+4b+b^{2}$$

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

$$8-6a-4b+b^{2}=12+4b+b^{2}$$

Subtract 8 from both sides.

$$-6a-4b+b^{2}=12+4b+b^{2}-8$$

Subtract 8 from 12 to get 4.

$$-6a-4b+b^{2}=4+4b+b^{2}$$

Add 4b to both sides.

$$-6a+b^{2}=4+4b+b^{2}+4b$$

Combine 4b and 4b to get 8b.

$$-6a+b^{2}=4+8b+b^{2}$$

Subtract b^{2} from both sides.

$$-6a=4+8b+b^{2}-b^{2}$$

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

$$-6a=4+8b$$

The equation is in standard form.

$$-6a=8b+4$$

Divide both sides by -6.

$$\frac{-6a}{-6}=\frac{8b+4}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$a=\frac{8b+4}{-6}$$

Divide 4+8b by -6.

$$a=\frac{-4b-2}{3}$$

Substitute \frac{-4b-2}{3} for a in the equation \sqrt{3+\left(1-a\right)^{2}+\left(2-b\right)^{2}}=\sqrt{4+\left(-2-a\right)^{2}+\left(-2-b\right)^{2}}.

$$\sqrt{3+\left(1-\frac{-4b-2}{3}\right)^{2}+\left(2-b\right)^{2}}=\sqrt{4+\left(-2-\frac{-4b-2}{3}\right)^{2}+\left(-2-b\right)^{2}}$$

Simplify. The value a=\frac{-4b-2}{3} satisfies the equation.

$$\frac{1}{3}\left(88+25b^{2}+4b\right)^{\frac{1}{2}}=\frac{1}{3}\left(88+25b^{2}+4b\right)^{\frac{1}{2}}$$

Equation \sqrt{\left(1-a\right)^{2}+\left(2-b\right)^{2}+3}=\sqrt{\left(-a-2\right)^{2}+\left(-b-2\right)^{2}+4} has a unique solution.

$$a=\frac{-4b-2}{3}$$

Solve for b

$$b=-\frac{3a}{4}-\frac{1}{2}$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3+\left(1-a\right)^{2}+\left(2-b\right)^{2}}\right)^{2}=\left(\sqrt{4+\left(-2-a\right)^{2}+\left(-2-b\right)^{2}}\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(1-a\right)^{2}.

$$\left(\sqrt{3+1-2a+a^{2}+\left(2-b\right)^{2}}\right)^{2}=\left(\sqrt{4+\left(-2-a\right)^{2}+\left(-2-b\right)^{2}}\right)^{2}$$

Add 3 and 1 to get 4.

$$\left(\sqrt{4-2a+a^{2}+\left(2-b\right)^{2}}\right)^{2}=\left(\sqrt{4+\left(-2-a\right)^{2}+\left(-2-b\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{4-2a+a^{2}+4-4b+b^{2}}\right)^{2}=\left(\sqrt{4+\left(-2-a\right)^{2}+\left(-2-b\right)^{2}}\right)^{2}$$

Add 4 and 4 to get 8.

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

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

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

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(-2-a\right)^{2}.

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

Add 4 and 4 to get 8.

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

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

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

Add 8 and 4 to get 12.

$$8-2a+a^{2}-4b+b^{2}=\left(\sqrt{12+4a+a^{2}+4b+b^{2}}\right)^{2}$$

Calculate \sqrt{12+4a+a^{2}+4b+b^{2}} to the power of 2 and get 12+4a+a^{2}+4b+b^{2}.

$$8-2a+a^{2}-4b+b^{2}=12+4a+a^{2}+4b+b^{2}$$

Subtract 4b from both sides.

$$8-2a+a^{2}-4b+b^{2}-4b=12+4a+a^{2}+b^{2}$$

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

$$8-2a+a^{2}-8b+b^{2}=12+4a+a^{2}+b^{2}$$

Subtract b^{2} from both sides.

$$8-2a+a^{2}-8b+b^{2}-b^{2}=12+4a+a^{2}$$

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

$$8-2a+a^{2}-8b=12+4a+a^{2}$$

Subtract 8 from both sides.

$$-2a+a^{2}-8b=12+4a+a^{2}-8$$

Subtract 8 from 12 to get 4.

$$-2a+a^{2}-8b=4+4a+a^{2}$$

Add 2a to both sides.

$$a^{2}-8b=4+4a+a^{2}+2a$$

Combine 4a and 2a to get 6a.

$$a^{2}-8b=4+6a+a^{2}$$

Subtract a^{2} from both sides.

$$-8b=4+6a+a^{2}-a^{2}$$

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

$$-8b=4+6a$$

The equation is in standard form.

$$-8b=6a+4$$

Divide both sides by -8.

$$\frac{-8b}{-8}=\frac{6a+4}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$b=\frac{6a+4}{-8}$$

Divide 4+6a by -8.

$$b=-\frac{3a}{4}-\frac{1}{2}$$

Substitute -\frac{3a}{4}-\frac{1}{2} for b in the equation \sqrt{3+\left(1-a\right)^{2}+\left(2-b\right)^{2}}=\sqrt{4+\left(-2-a\right)^{2}+\left(-2-b\right)^{2}}.

$$\sqrt{3+\left(1-a\right)^{2}+\left(2-\left(-\frac{3a}{4}-\frac{1}{2}\right)\right)^{2}}=\sqrt{4+\left(-2-a\right)^{2}+\left(-2-\left(-\frac{3a}{4}-\frac{1}{2}\right)\right)^{2}}$$

Simplify. The value b=-\frac{3a}{4}-\frac{1}{2} satisfies the equation.

$$\frac{1}{4}\left(164+25a^{2}+28a\right)^{\frac{1}{2}}=\frac{1}{4}\left(164+25a^{2}+28a\right)^{\frac{1}{2}}$$

Equation \sqrt{\left(1-a\right)^{2}+\left(2-b\right)^{2}+3}=\sqrt{\left(-a-2\right)^{2}+\left(-b-2\right)^{2}+4} has a unique solution.

$$b=-\frac{3a}{4}-\frac{1}{2}$$