| 答案 |
| 送交者: 車五進二 2019月10月24日23:54:08 於 [靈機一動] 發送悄悄話 |
| 回 答: 趣味的數學-117 由 gugeren 於 2019-10-21 19:02:41 |
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The intersection points between line y = kx and circle (x-3)^2 + (y-3)^2 = 6 is the roots to this system of equations, therefore (x-3)^2 + (kx-3)^2 = 6. Simplify yields (k^2+1)x^2 - (6k+6)x + 12 = 0, the line is tangent when there is only one solution for x to this quadratic equation, hence (6k+6)^2 - 4*12*(k^2+1) = 0. Solve for k = 3 + 2*sqrt(2) |
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