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
