Solve (x-a/2)^2+y^2=(a/2)^2

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

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{y^{2}}{x}+x\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&x=0\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

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

Express 2\left(-\frac{a}{2}\right) as a single fraction.

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

Cancel out 2 and 2.

$$x^{2}-ax+\left(-\frac{a}{2}\right)^{2}+y^{2}=\left(\frac{a}{2}\right)^{2}$$

Calculate -\frac{a}{2} to the power of 2 and get \left(\frac{a}{2}\right)^{2}.

$$x^{2}-ax+\left(\frac{a}{2}\right)^{2}+y^{2}=\left(\frac{a}{2}\right)^{2}$$

To raise \frac{a}{2} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}-ax+\frac{a^{2}}{2^{2}}+y^{2}=\left(\frac{a}{2}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-ax+y^{2} times \frac{2^{2}}{2^{2}}.

$$\frac{\left(x^{2}-ax+y^{2}\right)\times 2^{2}}{2^{2}}+\frac{a^{2}}{2^{2}}=\left(\frac{a}{2}\right)^{2}$$

Since \frac{\left(x^{2}-ax+y^{2}\right)\times 2^{2}}{2^{2}} and \frac{a^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}-ax+y^{2}\right)\times 2^{2}+a^{2}}{2^{2}}=\left(\frac{a}{2}\right)^{2}$$

Do the multiplications in \left(x^{2}-ax+y^{2}\right)\times 2^{2}+a^{2}.

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

To raise \frac{a}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{4x^{2}-4xa+4y^{2}+a^{2}}{2^{2}}=\frac{a^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4x^{2}-4xa+4y^{2}+a^{2}}{4}=\frac{a^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4x^{2}-4xa+4y^{2}+a^{2}}{4}=\frac{a^{2}}{4}$$

Subtract \frac{a^{2}}{4} from both sides.

$$\frac{4x^{2}-4xa+4y^{2}+a^{2}}{4}-\frac{a^{2}}{4}=0$$

Since \frac{4x^{2}-4xa+4y^{2}+a^{2}}{4} and \frac{a^{2}}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4x^{2}-4xa+4y^{2}+a^{2}-a^{2}}{4}=0$$

Combine like terms in 4x^{2}-4xa+4y^{2}+a^{2}-a^{2}.

$$\frac{-4xa+4x^{2}+4y^{2}}{4}=0$$

Divide each term of -4xa+4x^{2}+4y^{2} by 4 to get -xa+x^{2}+y^{2}.

$$-xa+x^{2}+y^{2}=0$$

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

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

Subtract y^{2} from both sides.

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

The equation is in standard form.

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

Divide both sides by -x.

$$\frac{\left(-x\right)a}{-x}=\frac{-x^{2}-y^{2}}{-x}$$

Dividing by -x undoes the multiplication by -x.

$$a=\frac{-x^{2}-y^{2}}{-x}$$

Divide -y^{2}-x^{2} by -x.

$$a=\frac{y^{2}}{x}+x$$

Solve for a

$$\left\{\begin{matrix}a=\frac{y^{2}}{x}+x\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&x=0\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

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

Express 2\left(-\frac{a}{2}\right) as a single fraction.

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

Cancel out 2 and 2.

$$x^{2}-ax+\left(-\frac{a}{2}\right)^{2}+y^{2}=\left(\frac{a}{2}\right)^{2}$$

Calculate -\frac{a}{2} to the power of 2 and get \left(\frac{a}{2}\right)^{2}.

$$x^{2}-ax+\left(\frac{a}{2}\right)^{2}+y^{2}=\left(\frac{a}{2}\right)^{2}$$

To raise \frac{a}{2} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}-ax+\frac{a^{2}}{2^{2}}+y^{2}=\left(\frac{a}{2}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-ax+y^{2} times \frac{2^{2}}{2^{2}}.

$$\frac{\left(x^{2}-ax+y^{2}\right)\times 2^{2}}{2^{2}}+\frac{a^{2}}{2^{2}}=\left(\frac{a}{2}\right)^{2}$$

Since \frac{\left(x^{2}-ax+y^{2}\right)\times 2^{2}}{2^{2}} and \frac{a^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}-ax+y^{2}\right)\times 2^{2}+a^{2}}{2^{2}}=\left(\frac{a}{2}\right)^{2}$$

Do the multiplications in \left(x^{2}-ax+y^{2}\right)\times 2^{2}+a^{2}.

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

To raise \frac{a}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{4x^{2}-4xa+4y^{2}+a^{2}}{2^{2}}=\frac{a^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4x^{2}-4xa+4y^{2}+a^{2}}{4}=\frac{a^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4x^{2}-4xa+4y^{2}+a^{2}}{4}=\frac{a^{2}}{4}$$

Subtract \frac{a^{2}}{4} from both sides.

$$\frac{4x^{2}-4xa+4y^{2}+a^{2}}{4}-\frac{a^{2}}{4}=0$$

Since \frac{4x^{2}-4xa+4y^{2}+a^{2}}{4} and \frac{a^{2}}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4x^{2}-4xa+4y^{2}+a^{2}-a^{2}}{4}=0$$

Combine like terms in 4x^{2}-4xa+4y^{2}+a^{2}-a^{2}.

$$\frac{-4xa+4x^{2}+4y^{2}}{4}=0$$

Divide each term of -4xa+4x^{2}+4y^{2} by 4 to get -xa+x^{2}+y^{2}.

$$-xa+x^{2}+y^{2}=0$$

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

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

Subtract y^{2} from both sides.

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

The equation is in standard form.

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

Divide both sides by -x.

$$\frac{\left(-x\right)a}{-x}=\frac{-x^{2}-y^{2}}{-x}$$

Dividing by -x undoes the multiplication by -x.

$$a=\frac{-x^{2}-y^{2}}{-x}$$

Divide -x^{2}-y^{2} by -x.

$$a=\frac{y^{2}}{x}+x$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{a^{2}-4y^{2}}+a}{2}$$

$$x=\frac{\sqrt{a^{2}-4y^{2}}+a}{2}$$

Solve for x

$$x=\frac{-\sqrt{a^{2}-4y^{2}}+a}{2}$$

$$x=\frac{\sqrt{a^{2}-4y^{2}}+a}{2}\text{, }|a|\geq 2|y|$$