show y=3x-5 is tangent to x^2 + y^2 +2x -4y - 5 = 0 and the point where they touch

y=3x-5x^2 + (3x-5)^2 + 2x - 4(3x-5) - 5 = 0x^2 + 9x^2 -30x +25 + 2x -12x + 20 - 5 = 010x^2 -40x + 40 = 010 (x^2 - 4x +4) = 010(x - 2)^2 = 0x=2implies one point of contact, therefore tangenty = 3x - 5y = 6 -5 = 1