Through P (4,1), make the tangent of the circle x ^ 2 + y ^ 2-6x-2y + 9 = 0, and find the tangent equation

Make the circle into the standard form: (x-3) ^ 2 + (Y-1) ^ 2 = 1
Then the general formula of the tangent equation of a fixed point (x0, Y0) passing through a circle is
(x0-3)(x-3)+(y0-1)(y-1)=1.
Substituting (4,1) gives x0 = 1, only this solution. Then the tangent is x = 4
If there are still two unknowns after substitution, the solution of circular equation is needed

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