A and b intersect to find the locus equation of AB midpoint.
[Solution] Let the linear equation be y=kx and substitute it into the cyclic equation.
(k^2+ 1)x^2-6x+5=0
From the relationship between root and coefficient, it is concluded that the abscissa of a and b satisfies
x 1+x2=6/(k^2+ 1)
Then substitute it into the linear equation.
, get a, b two ordinate satisfy.
y 1+y2=6k/(k^2+ 1)
The coordinates of the midpoint between a and b are
x=(x 1+x2)/2
y=(y 1+y2)/2
therefore
x=3/(k^2+ 1)
y=3k/(k^2+ 1)
Remove from the two formulas
The ballistic equation is obtained.
x^2+y^2-3x=0
solve an equation
x^2+y^2-3x=0
x^2+y^2-6x+5=0
get
x=5/3
y= 2√5/3
It can be determined that the trajectory is a following arc.
x^2+y^2-3x=0
x≥5/3
-2√5/3≤y≤2√5/3