equation of locus

Let B = (x1, y1), C = (x2, y2)
Let m be the slope of the line. Then the equation of the line is
y = m(x - 2)........(1)
x^2 + y^2 = 25..(2)
Put (1) into (2),
x^2 + m^2(x - 2)^2 = 25
x^2 + m^2 x^2 - 4m^2 x + 4m^2 - 25 = 0
(m^2 + 1)x^2 - 4m^2 x + (4m^2 - 25) = 0.............(*)
Note that x1, x2 are roots of (*).
x1 + x2 = 4m^2 / (m^2 + 1), x1x2 = (4m^2 - 25) / (m^2 + 1)
Let M = (x, y), then M lies on the line (1).
x = (1 / 2)(x1 + x2) = (1 / 2)[4m^2 / (m^2 + 1)] = 2m^2 / (m^2 + 1)..........(3)
y = m(x - 2) = m[2m^2 / (m^2 + 1) - 2] = m[2m^2 - 2m^2 - 2] / (m^2 + 1)
y = -2m / (m^2 + 1)............(4)
(3) / (4), x / y = 2m^2 / (-2m) = -m
m = -x / y.........(5)
Put (5) into (4),
y = -2[-x / y] / ([-x / y]^2 + 1)
y = 2(x / y) / [(x^2 + y^2) / y^2]
y = (2x / y)[y^2 / (x^2 + y^2)]
y = 2xy / (x^2 + y^2)
x^2 + y^2 = 2x
x^2 + y^2 - 2x = 0

let x, y be the point on the locus.
(x, y) is the mid-point of BC

let (a, b) be the point on the circle
(a1, b1) be B, (a2, b2) be C

x = (a1 + a2)/2 and y = (b1 + b2)/2
2x = a1 + a2 and 2y = b1 + b2
a^2 + b^2 = 25
b^2 = 25 - a^2

A, B, C and M are on a striaght line.
(y - 0)/(x - 2) = (b1 - b2)/(a1 - a2)
y/(x - 2) = (b1-b2)/(a1-a2) * (b1+b2)/(b1+b2) * (a1+a2)/(a1+a2)
y/(x - 2) = (b1^2-b2^2)/(a1^2-a2^2) * (a1+a1)/(b1+b2)
y/(x - 2) = (25-a1^2 - 25+a2^2)/(a1^2-a2^2) * (a1+a2)/(b1+b2)
y/(x - 2) = -1* 2x/2y
y/(x - 2)= -x/y
y^2 = 2x - x^2
x^2 + y^2 -2x = 0

the equation of the locus is x^2 + y^2 -2x = 0