find the equation of the tangent at (0,2) to the circle with
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find the equation of the tangent at (0,2) to the circle with equation
(x+2)平方 + (y+1)平方=13
(x+2)平方 + (y+1)平方=13
(x+2)平方 + (y+1)平方=13的圆心为(-2,-1),所以过圆心和(0,2)的直线斜率为:[2-(-1)]/(0-(-2))=3/2,所以切线方程显然与过圆心和(0,2)的直线斜率互为负倒数:所以为:-1/(3/2)=-2/3,已知斜率和过(0,2)点的条件,可得:y=(-2/3)x+2,整理得:2x+3y=6
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