locus problem

The center of the circle x² + y² = 25 is O(0,0). OM ⊥ AM (line joining centre to mid-pt. of chord⊥chord)Therefore Slope of OM * Slope of AM = - 1 . Let (x , y) be M , the locus of M is :(y - 0)/(x - 0) * (y - 0)/(x - 2) = - 1y² = - x(x - 2)x² + y² - 2x = 0
Notes : 1): It follows that the locus of M is the same as the circle on OA as diameter.2): The locus of M is independent of the radius of the circle x² + y² = 25.

Let B = (x1, y1), C = (x2, y2)

Let m be the slope of the line. Then the equation of the line isy = m(x - 2)........(1)x^2 + y^2 = 25..(2)Put (1) into (2), x^2 + m^2(x - 2)^2 = 25x^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) = -mm = -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 = 2xx^2 + y^2 - 2x = 0