Solve {l}{y-4=k(x-2)}{x^2+y^2=4}

$$\left. \begin{array} { l } { y - 4 = k ( x - 2 ) } \\ { x ^ { 2 } + y ^ { 2 } = 4 } \end{array} \right.$$

Solve for x, y

$$x=\frac{2\left(k^{2}-2k-\sqrt{4k-3}\right)}{k^{2}+1}\text{, }y=\frac{2\left(-\sqrt{4k-3}k-k+2\right)}{k^{2}+1}$$

$$x=\frac{2\left(k^{2}-2k+\sqrt{4k-3}\right)}{k^{2}+1}\text{, }y=\frac{2\left(\sqrt{4k-3}k-k+2\right)}{k^{2}+1}\text{, }k\geq \frac{3}{4}$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Use the distributive property to multiply k by x-2.

$$y-4=kx-2k$$

Subtract kx from both sides.

$$y-4-kx=-2k$$

Add 4 to both sides.

$$y-kx=-2k+4$$

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.

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

Solve y+\left(-k\right)x=4-2k for y by isolating y on the left hand side of the equal sign.

$$y+\left(-k\right)x=4-2k$$

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

$$y=kx+4-2k$$

Substitute kx+4-2k for y in the other equation, x^{2}+y^{2}=4.

$$x^{2}+\left(kx+4-2k\right)^{2}=4$$

Square kx+4-2k.

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

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

$$\left(k^{2}+1\right)x^{2}+2k\left(4-2k\right)x+\left(4-2k\right)^{2}=4$$

Subtract 4 from both sides of the equation.

$$\left(k^{2}+1\right)x^{2}+2k\left(4-2k\right)x+\left(4-2k\right)^{2}-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1k^{2} for a, 1\times 2k\left(4-2k\right) for b, and 12-16k+4k^{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2k\left(4-2k\right)±\sqrt{\left(2k\left(4-2k\right)\right)^{2}-4\left(k^{2}+1\right)\left(4k^{2}-16k+12\right)}}{2\left(k^{2}+1\right)}$$

Square 1\times 2k\left(4-2k\right).

$$x=\frac{-2k\left(4-2k\right)±\sqrt{16k^{2}\left(2-k\right)^{2}-4\left(k^{2}+1\right)\left(4k^{2}-16k+12\right)}}{2\left(k^{2}+1\right)}$$

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

$$x=\frac{-2k\left(4-2k\right)±\sqrt{16k^{2}\left(2-k\right)^{2}+\left(-4k^{2}-4\right)\left(4k^{2}-16k+12\right)}}{2\left(k^{2}+1\right)}$$

Multiply -4-4k^{2} times 12-16k+4k^{2}.

$$x=\frac{-2k\left(4-2k\right)±\sqrt{16k^{2}\left(2-k\right)^{2}-16\left(1-k\right)\left(3-k\right)\left(k^{2}+1\right)}}{2\left(k^{2}+1\right)}$$

Add 16k^{2}\left(2-k\right)^{2} to -16\left(1+k^{2}\right)\left(3-k\right)\left(1-k\right).

$$x=\frac{-2k\left(4-2k\right)±\sqrt{64k-48}}{2\left(k^{2}+1\right)}$$

Take the square root of -48+64k.

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

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

$$x=\frac{-4k\left(2-k\right)±4\sqrt{4k-3}}{2k^{2}+2}$$

Now solve the equation x=\frac{-4k\left(2-k\right)±4\sqrt{4k-3}}{2k^{2}+2} when ± is plus. Add -4k\left(2-k\right) to 4\sqrt{-3+4k}.

$$x=\frac{4k^{2}-8k+4\sqrt{4k-3}}{2k^{2}+2}$$

Divide -8k+4k^{2}+4\sqrt{-3+4k} by 2+2k^{2}.

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

Now solve the equation x=\frac{-4k\left(2-k\right)±4\sqrt{4k-3}}{2k^{2}+2} when ± is minus. Subtract 4\sqrt{-3+4k} from -4k\left(2-k\right).

$$x=\frac{4k^{2}-8k-4\sqrt{4k-3}}{2k^{2}+2}$$

Divide -8k+4k^{2}-4\sqrt{-3+4k} by 2+2k^{2}.

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

There are two solutions for x: \frac{2\left(-2k+k^{2}+\sqrt{-3+4k}\right)}{1+k^{2}} and \frac{2\left(-2k+k^{2}-\sqrt{-3+4k}\right)}{1+k^{2}}. Substitute \frac{2\left(-2k+k^{2}+\sqrt{-3+4k}\right)}{1+k^{2}} for x in the equation y=kx+4-2k to find the corresponding solution for y that satisfies both equations.

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

Multiply k times \frac{2\left(-2k+k^{2}+\sqrt{-3+4k}\right)}{1+k^{2}}.

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

Add k\times \frac{2\left(-2k+k^{2}+\sqrt{-3+4k}\right)}{1+k^{2}} to 4-2k.

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

Now substitute \frac{2\left(-2k+k^{2}-\sqrt{-3+4k}\right)}{1+k^{2}} for x in the equation y=kx+4-2k and solve to find the corresponding solution for y that satisfies both equations.

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

Multiply k times \frac{2\left(-2k+k^{2}-\sqrt{-3+4k}\right)}{1+k^{2}}.

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

Add k\times \frac{2\left(-2k+k^{2}-\sqrt{-3+4k}\right)}{1+k^{2}} to 4-2k.

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

The system is now solved.

$$y=\frac{2\left(k^{2}-2k+\sqrt{4k-3}\right)}{k^{2}+1}k-2k+4,x=\frac{2\left(k^{2}-2k+\sqrt{4k-3}\right)}{k^{2}+1}\text{ or }y=\frac{2\left(k^{2}-2k-\sqrt{4k-3}\right)}{k^{2}+1}k-2k+4,x=\frac{2\left(k^{2}-2k-\sqrt{4k-3}\right)}{k^{2}+1}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{2\left(k^{2}-2k-\sqrt{4k-3}\right)}{k^{2}+1}\text{, }y=\frac{2\left(-\sqrt{4k-3}k-k+2\right)}{k^{2}+1}\text{; }x=\frac{2\left(k^{2}-2k+\sqrt{4k-3}\right)}{k^{2}+1}\text{, }y=\frac{2\left(\sqrt{4k-3}k-k+2\right)}{k^{2}+1}\text{, }&k\neq -i\text{ and }k\neq i\\x=-\frac{\left(1-k\right)\left(3-k\right)}{k\left(2-k\right)}\text{, }y=\frac{k^{2}-4k+5}{2-k}\text{, }&k=-i\text{ or }k=i\end{matrix}\right.$$