"A circle passes through the points A, B, and C which have coordinates respectively. Equation of circle

A line y=mx-3of variable gradient m, cuts the circle at L and M. Find the Cartesian equation of the locus of the midpoint of the line LM."

How i do it . . . .

x^2 + y^2 - 2y = 3 - - - - (1)

and y = mx + 3 - - - - (2) , substituting

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x^2 + (mx + 3)^2 - 2(mx + 3) = 3,

x^2 + m^2x^2 + 6mx + 9 - 2mx - 6 = 3,

(1 + m^2)x^2 + 4mx = 3 - 9 + 6 = 0,

(1 + m^2)x^2 + 4mx = 0

x[(1 + m^2)x + 4m] = 0,

x = 0, then y = 3

x = -4m/(1 + m^2), then y = (3 - m^2)/(1 + m^2)

i don't know what to continue either . . ..