Let the centre of the circle be P(x,y) and the radius be r.
A(1,-6) B(7,0) C(-1,4) and D(a,2) are equidistant from P.
Therefore, AP2=CP2
By distance formula;
(x-1)2 + (y+6)2=(x+1)2+(y-4)2
This gives x-5y-5=0 -------(i)
Also AP2=BP2:
(x-1)2 + (y+6)2 = (x-7)2 + y2
This gives: x+y-1=0------(ii)
Subtracting i from ii
We get 6y+4=0 or y=-2/3
Putting this in (ii)
We get x=5/3
Now, AP2= DP2
(x-1)2 + (y+6)2 = (x-a)2 +(y-2)2
Putting values of x and y
We get: 3a2-10a-57=0
Applying quadratic formula, we get a= 19/3 or -3.