From the point P (1, - 2) to the circle x ^ 2 + y ^ 2-6x-2y + 6 = 0, the tangent equation is
Suppose the tangent equation
y+2=k(x-1)
y=kx-k-2
kx-y-k-2=0
x^2+y^2-6x-2y+6=0
(x-3)^2+(y-1)^2=4
Center (3,1) radius 2
Distance from center to tangent = 2
2 = | 3k-1-k-2 | / root (k ^ 2 + 1)
k=5/12
The tangent equation is
y=5(x-1)/12-2
Obviously, x = 1 is also tangent
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