若直线l过点P(2,3),且与圆(X-1)2+(Y-1)2=1相切的切线方程
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y-3=k(x-2)
kx-y-2k+3=0
(1,1),R=1
/k-1-2k+3//(k^2+1)^1/2=1
/-k+2/=(k^2+1)^1/2
k^2-4k+4=k^2+1
4k=3
k=3/4
y-3=3/4(x-2)
4y-12=3x-6
3x-4y+6=0
kx-y-2k+3=0
(1,1),R=1
/k-1-2k+3//(k^2+1)^1/2=1
/-k+2/=(k^2+1)^1/2
k^2-4k+4=k^2+1
4k=3
k=3/4
y-3=3/4(x-2)
4y-12=3x-6
3x-4y+6=0
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圆心(1,1),设切点为(a,b)
(b - 1)² + (a - 1)² = 1 => a² + b² - 2a - 2b + 1 = 0
圆心到切点的切线的斜率 = (b - 1)/(a - 1)
切线斜率 = -1/[(b - 1)/(a - 1)] = (1 - a)/(b - a)
由(a,b)和(2,3)得切线斜率 = (b - 3)/(a - 2)
∴(1 - a)/(b - 1) = (b - 3)/(a - 2) => a² + b² - 3a - 4b + 5 = 0
{ a² + b² - 2a - 2b + 1 = 0
{ a² + b² - 3a - 4b + 5 = 0
解方程得:
{ a = 2/5,b = 9/5
{ a = 2,b = 1
切线方程为(y - 9/5)/(x - 2/5) = (3 - 9/5)/(2 - 2/5) => y = (3/4)x + 3/2
(y - 1)/(x - 2) = (3 - 1)/(2 - 2)
(y - 1)(2 - 2) = 2(x - 2)
2(x - 2) = 0
x = 2
所以切线方程分别为y = (3/4)x + 3/2 或 x = 2
(b - 1)² + (a - 1)² = 1 => a² + b² - 2a - 2b + 1 = 0
圆心到切点的切线的斜率 = (b - 1)/(a - 1)
切线斜率 = -1/[(b - 1)/(a - 1)] = (1 - a)/(b - a)
由(a,b)和(2,3)得切线斜率 = (b - 3)/(a - 2)
∴(1 - a)/(b - 1) = (b - 3)/(a - 2) => a² + b² - 3a - 4b + 5 = 0
{ a² + b² - 2a - 2b + 1 = 0
{ a² + b² - 3a - 4b + 5 = 0
解方程得:
{ a = 2/5,b = 9/5
{ a = 2,b = 1
切线方程为(y - 9/5)/(x - 2/5) = (3 - 9/5)/(2 - 2/5) => y = (3/4)x + 3/2
(y - 1)/(x - 2) = (3 - 1)/(2 - 2)
(y - 1)(2 - 2) = 2(x - 2)
2(x - 2) = 0
x = 2
所以切线方程分别为y = (3/4)x + 3/2 或 x = 2
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