Solve {c}{(x-2)^2+(y-1)^2=1}{y=kx+b}

$$\left. \begin{array} { c } { ( x - 2 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 1 } \\ { y = k x + b } \end{array} \right.$$

Solve for x, y (complex solution)

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

Solve for x, y

$$x=\frac{-bk-\sqrt{4k-3k^{2}+2b-4bk-b^{2}}+k+2}{k^{2}+1}\text{, }y=\frac{-k\sqrt{4k-3k^{2}+2b-4bk-b^{2}}+k^{2}+b+2k}{k^{2}+1}$$

$$x=\frac{-bk+\sqrt{4k-3k^{2}+2b-4bk-b^{2}}+k+2}{k^{2}+1}\text{, }y=\frac{k\sqrt{4k-3k^{2}+2b-4bk-b^{2}}+k^{2}+b+2k}{k^{2}+1}\text{, }k\geq \frac{-\sqrt{b^{2}-2b+4}-2b+2}{3}\text{ and }k\leq \frac{\sqrt{b^{2}-2b+4}-2b+2}{3}$$