Solve {l}{x^2+y^2=R^2}{y=mx}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = R ^ { 2 } } \\ { y = m x } \end{array} \right.$$

Solve for x, y

$$x=-\frac{R}{\sqrt{m^{2}+1}}\text{, }y=-\frac{Rm}{\sqrt{m^{2}+1}}$$

$$x=\frac{R}{\sqrt{m^{2}+1}}\text{, }y=\frac{Rm}{\sqrt{m^{2}+1}}$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract mx from both sides.

$$y-mx=0$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$y+\left(-m\right)x=0,x^{2}+y^{2}=R^{2}$$

Solve y+\left(-m\right)x=0 for y by isolating y on the left hand side of the equal sign.

$$y+\left(-m\right)x=0$$

Subtract \left(-m\right)x from both sides of the equation.

$$y=mx$$

Substitute mx for y in the other equation, x^{2}+y^{2}=R^{2}.

$$x^{2}+\left(mx\right)^{2}=R^{2}$$

Square mx.

$$x^{2}+m^{2}x^{2}=R^{2}$$

Add x^{2} to m^{2}x^{2}.

$$\left(m^{2}+1\right)x^{2}=R^{2}$$

Subtract R^{2} from both sides of the equation.

$$\left(m^{2}+1\right)x^{2}-R^{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1m^{2} for a, 1\times 0\times 2m for b, and -R^{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(m^{2}+1\right)\left(-R^{2}\right)}}{2\left(m^{2}+1\right)}$$

Square 1\times 0\times 2m.

$$x=\frac{0±\sqrt{-4\left(m^{2}+1\right)\left(-R^{2}\right)}}{2\left(m^{2}+1\right)}$$

Multiply -4 times 1+1m^{2}.

$$x=\frac{0±\sqrt{\left(-4m^{2}-4\right)\left(-R^{2}\right)}}{2\left(m^{2}+1\right)}$$

Multiply -4-4m^{2} times -R^{2}.

$$x=\frac{0±\sqrt{4R^{2}\left(m^{2}+1\right)}}{2\left(m^{2}+1\right)}$$

Take the square root of 4\left(1+m^{2}\right)R^{2}.

$$x=\frac{0±2|R|\sqrt{m^{2}+1}}{2\left(m^{2}+1\right)}$$

Multiply 2 times 1+1m^{2}.

$$x=\frac{0±2|R|\sqrt{m^{2}+1}}{2m^{2}+2}$$

Now solve the equation x=\frac{0±2|R|\sqrt{m^{2}+1}}{2m^{2}+2} when ± is plus.

$$x=\frac{|R|}{\sqrt{m^{2}+1}}$$

Now solve the equation x=\frac{0±2|R|\sqrt{m^{2}+1}}{2m^{2}+2} when ± is minus.

$$x=-\frac{|R|}{\sqrt{m^{2}+1}}$$

There are two solutions for x: \frac{|R|}{\sqrt{1+m^{2}}} and -\frac{|R|}{\sqrt{1+m^{2}}}. Substitute \frac{|R|}{\sqrt{1+m^{2}}} for x in the equation y=mx to find the corresponding solution for y that satisfies both equations.

$$y=m\times \frac{|R|}{\sqrt{m^{2}+1}}$$

Multiply m times \frac{|R|}{\sqrt{1+m^{2}}}.

$$y=\frac{|R|}{\sqrt{m^{2}+1}}m$$

Now substitute -\frac{|R|}{\sqrt{1+m^{2}}} for x in the equation y=mx and solve to find the corresponding solution for y that satisfies both equations.

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

Multiply m times -\frac{|R|}{\sqrt{1+m^{2}}}.

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

The system is now solved.

$$y=\frac{|R|}{\sqrt{m^{2}+1}}m,x=\frac{|R|}{\sqrt{m^{2}+1}}\text{ or }y=\left(-\frac{|R|}{\sqrt{m^{2}+1}}\right)m,x=-\frac{|R|}{\sqrt{m^{2}+1}}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\left(m^{2}+1\right)^{-\frac{1}{2}}R\text{, }y=\left(m^{2}+1\right)^{-\frac{1}{2}}Rm\text{; }x=-\left(m^{2}+1\right)^{-\frac{1}{2}}R\text{, }y=-\left(m^{2}+1\right)^{-\frac{1}{2}}Rm\text{, }&m\neq -i\text{ and }m\neq i\\x\in \mathrm{C}\text{, }y=mx\text{, }&\left(m=i\text{ or }m=-i\right)\text{ and }R=0\end{matrix}\right.$$