(x-2k)^2+(y-k)^2=1 шешу

$$( x - 2 k ) ^ { 2 } + ( y - k ) ^ { 2 } = 1$$

k мәнін табыңыз (complex solution)

$$k=\frac{\sqrt{5-4y^{2}+4xy-x^{2}}+y+2x}{5}$$

$$k=\frac{-\sqrt{5-4y^{2}+4xy-x^{2}}+y+2x}{5}$$

x мәнін табыңыз (complex solution)

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

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

k мәнін табыңыз

$$k=\frac{\sqrt{5-4y^{2}+4xy-x^{2}}+y+2x}{5}$$

$$k=\frac{-\sqrt{5-4y^{2}+4xy-x^{2}}+y+2x}{5}\text{, }x\geq 2y-\sqrt{5}\text{ and }x\leq 2y+\sqrt{5}$$

x мәнін табыңыз

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

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