Question 969254
Slope of line (or say chord) joining the two points on circumference 
= {{{(y2-y1)/(x2-x1)}}}
= {{{(-7-9)/(1-5)}}}
= {{{4}}}
So, slope of perpendicular bisector of this chord 
={{{-1/4}}} (For perpendicular lines, product of slopes = -1)
So let the equation of perpendicular bisector of this chord(which shall pass through center of circle) be given by
{{{y= (-1/4)x+c}}} -(i) (slope-intercept form of a line)
Now mid point of the chord is given by
{{{((1+5)/2)}}},{{{((9-7)/2)}}}
i.e.(3,1)
Since this should lie on the line given by eqn (i)above,
So, {{{1=(-1/4)*3+c}}}
i.e. {{{c=7/4}}}
So eqn of perpendicular bisector passing through center becomes
{{{y= (-1/4)x+7/4}}}
or, {{{x=7-4y}}}
Taking center as (7-4y,y), Its distance from both circumferential points is given by radii 
So, for point (5,9)
{{{(7-4y-5)^2+(y-9)^2 = 85}}} (Using Distance formula)
i.e. {{{17y^2+34y =0}}}
i.e. {{{y=0 or 2}}}
So, {{{x= 7 or -1}}}
There will be two required circles with centers at (7,0) and (-1,2) and radius sqrt(85)
So, the equation of cirles be given by:
{{{(x-7)^2+(y-0)^2=85}}} and {{{(x+1)^2+(y-2)^2=85}}}
Simplify, if you want.