Question

Выразить х через у. $y=2+\sqrt{2x-x^{2} }$

Answer 1

$y = 2 + \sqrt{2x - {x}^{2} } \\ \sqrt{2x - {x}^{2} } = y - 2 \\ 2x - {x}^{2} = (y - 2)^{2} \\ {x}^{2} - 2x + (y - 2)^{2} = 0 \\ D = {2}^{2} - 4(y - 2)^{2} = 4 - 4( {y}^{2} - 4y + 4) = \\ = 4 - 4 {y}^{2} + 16y - 16 = - 4 {y}^{2} + 16y - 12 = \\ = - 4( {y}^{2} - 4y + 3) = - 4(y - 1)(y - 3) \\ x_{1} = \frac{2 - \sqrt{ - 4(y - 1)(y - 3)} }{2} = 1 - \sqrt{(1 - y)(y - 3)} \\ x_{2} = \frac{2 + \sqrt{ - 4(y - 1)(y - 3)} }{2} = 1 + \sqrt{(1 - y)(y - 3)}$