Question

Не выполняя построения найдите координаты точек пересечения окружности x^2+y^2=17 и прямой 5х-3у=17

Answer 1

$x^2 + y^2 = 17, \ 5x - 3y = 17\\\\ 3y = 5x - 17\\\\ y = \frac{5}{3}x - \frac{17}{3}\\\\ x^2 + (\frac{5}{3}x - \frac{17}{3})^2 = 17\\\\ x^2 + \frac{25}{9}x^2 - \frac{170}{9}x + \frac{289}{9} = 17$

$\frac{34}{9}x^2 - \frac{170}{9}x + \frac{289}{9} = 17 \ | \ * \frac{9}{17}\\\\ 2x^2 - 10x + 17 = 9\\\\ 2x^2 - 10x + 8 = 0 \ | \ : 2\\\\ x^2 - 5x + 4 = 0\\\\ x^2 - 4x - x + 4 = 0\\\\ x(x - 4) - (x - 4) = 0\\\\ (x - 1)(x - 4) = 0\\\\ x_1 = 1, \ x_2 = 4\\\\ y_1 = -4, \ y_2 = 1$