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(x-1)²;+(y-2)²;=4的圓心為(1,2)半徑R=2 設過p(3,1)的直線方程為y-1=K(x-3) 即kx-y+1-3k=0 圓心到直線的距離 :|k-2+1-3k|/√k²;+(-1)²;=2 ∴k=3/4 所以這條切線為y-1=(3/4)(x-3)3 即3X-4y-5=0 由於X=3與圓相切 故過p(3,1)的另一條切線方程為X=3
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