Equation of a circle with center at#(h,k)# and a radius #r# is given by : #(x-h)^2+(y-k)^2=r^2#
As #AB# is the diameter of the circle, the midpoint #O# of #AB# is the center of the circle,
#=># midpoint #O=((1+9)/2,(5+1)/2)=(5,3)#
#=> r^2=(9-5)^2+(1-3)^2=20#
Hence, equation of the circle is :#(x-5)^2+(y-3)^2=20#
Solution 2) :
Let #P# be any point on the circle,
as #AB# is the diameter, #=> angleAPB=90^@#
let #m_1, m_2# be the slopes of #AP and BP#, respectively,
#=> m_1=(y-5)/(x-1)#,
#=> m_2=(y-1)/(x-9)#,
since #AP and BP# are perpendicular to each other, their products of the slopes is #-1, => m_1xxm_2=-1#,
#=> (y-5)/(x-1)*(y-1)/(x-9)=-1#
#=> (x-1)(x-9)=-(y-5)(y-1)#
#=> (x-1)(x-9)+(y-5)(y-1)=0#
#=> x^2-10x+y^2-6y=-14#
#=> (x-5)^2-25+(y-3)^2-9=-14#
#=> (x-5)^2+(y-3)^2=-14+25+9#
#=> (x-5)^2+(y-3)^2=20#
