42
правки
Изменения
Нет описания правки
(\alpha\bar{a}+\beta\bar{b}-\bar{a})^2=r_1^2\\
\end{array}
\right.</tex><br><tex>\left\{\begin{array}{lrl}\alpha^2\bar{a}^2+\beta^2\bar{b}^2+2\alpha\beta\bar{a}\bar{b}=r_0^2\ \ \ (1)\\((\alpha-1)\bar{a}+\beta\bar{b})^2=r_1^2\\\end{array}\right.</tex><br>Заметим, что в уравнении <tex>(1)</tex> третье слагаемое в правой части равно <tex>0</tex>, т.к. векторы <tex>\bar{a}</tex> и <tex>\bar{b}</tex> перпендикулярны.<br><tex>\left\{\begin{array}{lrl}\alpha^2\bar{a}^2+\beta^2\bar{b}^2=r_0^2\\(\alpha-1)^2\bar{a}^2+\beta^2\bar{b}^2=r_1^2\\\end{array}\right.</tex><br><tex>\left\{\begin{array}{lrl}2\alpha-1=\frac{r_0^2-r_1^2}{\bar{a}^2}\\\beta^2\bar{b}^2=r_0^2-\alpha^2\bar{a}^2\\\end{array}\right.</tex><br><tex>\left\{\begin{array}{lrl}\alpha=\frac{r_0^2-r_1^2+\bar{a}^2}{2\bar{a}^2}\\\beta^2\bar{b}^2=r_0^2-\alpha^2\bar{a}^2\\\end{array}\right.</tex><br><tex>\left\{\begin{array}{lrl}\alpha=\frac{r_0^2-r_1^2+\bar{a}^2}{2\bar{a}^2}\\\beta^2=\frac{r_0^2-\frac{(r_0^2-r_1^2+\bar{a}^2)^2}{4\bar{a}^4}\bar{a}^2}{\bar{b}^2}\\\end{array}\right.</tex><br><tex>\left\{\begin{array}{lrl}\alpha=\frac{r_0^2-r_1^2+\bar{a}^2}{2\bar{a}^2}\\\beta=\pm\frac{1}{|\bar{a}||\bar{b}|}\sqrt{-\frac{1}{2}(r_0^4+r_1^4+\bar{a}^4-2r_0^2r_1^2-2r_1^2\bar{a}^2)}\\\end{array}\right.</tex><br>
