find the equation of the tangent at (0,2) to the circle with equation(x+2)平方 + (y+1)平方=13

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findtheequationofthetangentat(0,2)tothecirclewithequation(x+2)平方+(y+1)平方=13findtheequationofthetange

find the equation of the tangent at (0,2) to the circle with equation(x+2)平方 + (y+1)平方=13
find the equation of the tangent at (0,2) to the circle with equation
(x+2)平方 + (y+1)平方=13

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

找到过点(0,2)的圆的切线方程
答案
2x+3y=6
因为该点在圆上,所以过这点只有一条切线
解法:
代入点可知:2*2+3*3=13
知点在圆上,所以这点既在圆上又在切线上
所以有
(2+0)*(2+x)+(2+1)*(y+1)=13
整理得:2x+3y=6
即切线方程...

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找到过点(0,2)的圆的切线方程
答案
2x+3y=6
因为该点在圆上,所以过这点只有一条切线
解法:
代入点可知:2*2+3*3=13
知点在圆上,所以这点既在圆上又在切线上
所以有
(2+0)*(2+x)+(2+1)*(y+1)=13
整理得:2x+3y=6
即切线方程

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题目意思为:求过圆上一点(0,2)的切线方程。
设此点为A(0,2),圆心O坐标为(-2,-1),则直线OA斜率k=(-1-2)/(-2)=3/2.
设所求切线L斜率为k′,由相切关系,得直线OA垂直于直线L,故有斜率之积k*k′=-1,
从而得k′=-2/3.
所求切线方程为:y-2=-2/3x,即 y=-2/3x+2....

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题目意思为:求过圆上一点(0,2)的切线方程。
设此点为A(0,2),圆心O坐标为(-2,-1),则直线OA斜率k=(-1-2)/(-2)=3/2.
设所求切线L斜率为k′,由相切关系,得直线OA垂直于直线L,故有斜率之积k*k′=-1,
从而得k′=-2/3.
所求切线方程为:y-2=-2/3x,即 y=-2/3x+2.

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2x+3y-6=0

依题求点(0,2)处的切线方程,设圆心为O,并有A(0,2)
易得直线OA的方程为y=3/2x+2,由于切线垂直于OA
则切线斜率为-2/3,即y=-2/3x+b,代入A(0,2)得b=2
即该圆O在点(0,2)处的切线方程为y=-2/3x+2

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