Find the linear equation parallel to the line x + y + 3 = 0 and tangent to the square of circle x + y-6x-4y + 5 = 0

X ^ 2 + y ^ 2-6x-4y + 5 = 0, (x-3) ^ 2 + (Y-2) ^ 2 = 8, the line parallel to x + y + 3 = 0 is set as x + y + a = 0, the distance from point (3,2) to the line is 8 ^ 1 / 2, 8 = | 3 + 2 + a | ^ 2 / 2, a = - 1 or a = - 9, the line is x + Y-1 = 0 or x + Y-9 = 0

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