Question

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

Answer 1

设直线方程是y-3=k(x-2)，即kx-y-2k+3=0
∵圆心是(1,1)，半径是1
又∵直线与圆相切，
∴|k-1-2k+3|/√(1+k²)=1
解得k=3/4，
∴切线方程是3/4x-y+3/2=0，即3x-4y+6=0
又∵x=2也过点（2,3），并且也与圆相切，所以x=2也是切线方程。
综上所述，切线方程是x=2和3x-4y+6=0

Answer 2

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

Answer 3

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