Parabola

Find the equation of the circle described on a focal chord with slope m, of the parabola y²=4ax as diameter.

There are two approaches to this problem.

1) The end points of the focal chord are (a/m²,2a/m) and (am²,-2am)

So the equation of the circle will be (x-a/m²)(x-am²) + (y-2a/m)(y+2am)=0

on solving
x² + y² -a(m² - 1/m²)x + 2a(m - 1/m)y -3a²=0

2) The slope of the line joining focus and one of the end points of focal chord (at²,2at) is

2at/at²-a = m
i.e. mt² - 2t - m = 0
from here, t₁ + t₂ = 2/m and t₁t₂ = -1
so equation of the circle will be (final answer)-

x² - a[4/m² +2]x + a² +y² - 4ay/m - 4a² = 0

I wanted if any approach is wrong (as final answers are different)
Thanks