Let #P(h,k)# be the center the circle.
Then, #k=h/4+4=(h+16)/4#....[1]
Also it passes through #(3,7)# and #(7,1)#
The equation of circle is given by
#rarr(x-h)^2+(y-k)^2=r^2# where #(x,y)# is passing point and #(h,k)# is center and #r# is the radius.
So, #rarr(3-h)^2+(7-k)^2=r^2#
#rarr9-6h+h^2+49-14k+k^2=r^2#
#rarrh^2+k^2-6h-14h+58=r^2#.....[2]
and #(7-h)^2+(1-k)^2=r^2#
#rarr49-14h+h^2+1-2k+k^2=r^2#
#rarrh^2+k^2-14h-2k+50=r^2#.....[3]
From [2 and [3], we have,
#rarrcancel(h^2+k^2)-6h-14h+58=cancel(h^2+k^2)-14h-2k+50#
#rarr8h-12k+8=0#
#rarr4*(2h-3k+2)=0#
#rarr2h-3k+2=0#....[4]
Putting the value of #k# from euation [1] to equation [4],
#rarr2h-3*(h+16)/4+2=0#
#rarr8h-3h-48+8=0#
#rarrh=8#
We have, #k=(h+16)/4#
#rarrk=(16+8)/4=6#
Now, #r^2=(7-h)^2+(1-k)^2=(7-8)^2+(1-6)^2=26#
The required equation is
#rarr(x-h)^2+(y-k)^2=r^2#
#rarr(x-8)^2+(y-6)^2=26#
#rarrx^2-16x+64+y^2-12y+36=26#
#rarrx^2+y^2-16x-12y+74=0#
