Given that the equation of circle is (x-1) + (Y-1) = 1, the coordinates of point P are (2,3), the tangent equation of point P is obtained urgent

Given that the equation of circle is (x-1) + (Y-1) = 1, the coordinates of point P are (2,3), the tangent equation of point P is obtained urgent

The coordinates of the center of the circle are o (1,1), and the radius of the circle is r = 1. Let the linear equation y = K (X-2) + 3 be changed into the general formula kx-y + (3-2k) = 0. Because the line is tangent to the circle, the distance from the center of the circle to the straight line is equal to the radius of the circle | k-1 + (3-2k) | / radical (K + (- 1)) = 1 (2-k) = K + 1, 4-4k + k = K + 1, k = 3 / 4. Because there should be two tangents meeting the requirements, the equations of these two tangents are: y = (3 / 4) (X-2) + 3, The general formula of x = 2 is: 3x-4y + 6 = 0, X-2 = 0