Challenge!

1
circle x^2+y^2-8x-2y+12=0
(x-4)^2+(y-1)^2=5
so center (4,1) radius √5
let the line whose midpoint is A intersect circle at BC
the slope of OA
=(1+5)/(4-3)
=6
so the slope of BC
=-1/6
Using point-slope form
the equation of the chord of the circle whose midpoint is A is
y+5=-1/6(x-3)
x+6y+27=0
2
Obviously, the shortest chord is obtained when it is prepandicular to the x-axis
the equation is x=3
the longest chord is obtained when it pass through the center
so
using two point form, the equation is
(y-0)/(x-3)=(1-0)/(4-3)
y=x-3
x-y-3=0