Find the tangent equation where the square of point P (3.1) and circle (x-1) plus the square of Y equals 4

The center of (x-1) &# 178; + (Y-2) &# 178; = 4 is (1,2) radius r = 2
Let the linear equation over P (3,1) be Y-1 = K (x-3)
That is kx-y + 1-3k = 0
Distance from center of circle to straight line
:|k-2+1-3k|/√k²+(-1)²=2
∴k=3/4
So this tangent is Y-1 = (3 / 4) (x-3) 3
That is, 3x-4y-5 = 0
Because x = 3 is tangent to the circle
So another tangent equation passing through P (3,1) is x = 3

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