{l}{y=kx+m}{(x-3)^2+y^2=6} ઉકેલો

$$\left\{ \begin{array} { l } { y = k x + m } \\ { ( x - 3 ) ^ { 2 } + y ^ { 2 } = 6 } \end{array} \right.$$

x, y માટે ઉકેલો (જટિલ સમાધાન)

$$\left\{\begin{matrix}x=\frac{-km-\sqrt{6-3k^{2}-6km-m^{2}}+3}{k^{2}+1}\text{, }y=\frac{-k\sqrt{6-3k^{2}-6km-m^{2}}+m+3k}{k^{2}+1}\text{; }x=\frac{-km+\sqrt{6-3k^{2}-6km-m^{2}}+3}{k^{2}+1}\text{, }y=\frac{k\sqrt{6-3k^{2}-6km-m^{2}}+m+3k}{k^{2}+1}\text{, }&k\neq -i\text{ and }k\neq i\\x=-\frac{m^{2}+3}{2\left(km-3\right)}\text{, }y=\frac{km^{2}-6m-3k}{2\left(km-3\right)}\text{, }&\left(m\neq -3i\text{ and }k=i\right)\text{ or }\left(m\neq 3i\text{ and }k=-i\right)\end{matrix}\right.$$

x, y માટે ઉકેલો

$$x=\frac{-km-\sqrt{6-3k^{2}-6km-m^{2}}+3}{k^{2}+1}\text{, }y=\frac{-k\sqrt{6-3k^{2}-6km-m^{2}}+m+3k}{k^{2}+1}$$

$$x=\frac{-km+\sqrt{6-3k^{2}-6km-m^{2}}+3}{k^{2}+1}\text{, }y=\frac{k\sqrt{6-3k^{2}-6km-m^{2}}+m+3k}{k^{2}+1}\text{, }k\geq -\frac{\sqrt{6m^{2}+18}}{3}-m\text{ and }k\leq \frac{\sqrt{6m^{2}+18}}{3}-m$$