find the locus of midpoints of the chords to the circle (x-3)^2+(y-2)^2=1 passing through the point (3,7)

Find the locus of midpoint of the chords to the circle (x-3)2+(y-2)2=1 passing through the point(3,7)

Let midpoint be (h,k) then we know that perpendicular from the centre to the chord is its perpendicular bisector so we get a line from centre ie. (3,2) to midpoint (h,k) which is perpendicular to the line from (h,k) to (3,7) as it lies on the chord. Use m1*m2=(-1) and get the required equation.