Solve a+sqrt{a^2+b^2}=2

$$a + \sqrt { a ^ { 2 } + b ^ { 2 } } = 2$$

Solve for a

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

Solution Steps

Subtract a from both sides of the equation.

$$\sqrt{a^{2}+b^{2}}=2-a$$

Square both sides of the equation.

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

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

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

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

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

Add 4a to both sides.

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

Subtract a^{2} from both sides.

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

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

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

Subtract b^{2} from both sides.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -b^{2}+4 by 4.

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

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

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

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

$$-\frac{1}{4}b^{2}+1+\frac{1}{4}\left(16+8b^{2}+b^{4}\right)^{\frac{1}{2}}=2$$

Equation \sqrt{a^{2}+b^{2}}=2-a has a unique solution.

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

Solve for b

$$b=2\sqrt{1-a}$$

$$b=-2\sqrt{1-a}\text{, }a\leq 1$$