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x²;+y²;= 4,圓心為原點(0,0),半徑為2的圓 直線x-my+2 = 0必定經過點(0,2),剛好在圓上, 那麼若: ①m=0,那麼x = -2,直線與圓相切與點(0,2) ②m≠0,那麼直線與圓必有多餘1個的交點,其實就是2個交點
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