amath

Let equation of the line be y = mx.
Sub. into equation of the circle we get
m^2x^2 + x^2 - 4x - 4 = 0
(1 + m^2)x^2 - 4x - 4 = 0
Roots of this equation gives the intersecting points of the line and the circle, that is x1 and x2.
b) Since sum of roots = x1 + x2 = 4/(1+m^2), so
x- coordinate of P is (x1 +x2)/2 = 2/(1 + m^2)..................(1)
y- coordinate of P = mx = 2m/(1 + m^2).......................(2)
Since y = mx, that is m = y/x, sub. into (1) we get
x = 2/(1 + y^2/x^2)
x = 2x^2/(x^2 + y^2)
x^2 + y^2 - 2x = 0 is the locus of P.