Here is an approach using linear regression:
circfit[pts_] := Module[{reg, lm, bf, exp, center, rad},
reg = {2 #1, 2 #2, #2^2 + #1^2} & @@@ pts;
lm = LinearModelFit[reg, {1, x, y}, {x, y}];
bf = lm["BestFitParameters"];
exp = (x - #2)^2 + (y - #3)^2 - #1 - #2^2 - #3^2 & @@ bf;
{center, rad} = {{#2, #3}, Sqrt[#2^2 + #3^2 + #1]} & @@ bf;
circlefit[{"expression" -> exp, "center" -> center,
"radius" -> rad}]]; circlefit[list_][field_] := field /. list;
circlefit[list_]["Properties"] := list /. Rule[field_, _] :> field;
circlefit /: ReplaceAll[fields_, circlefit[list_]] := fields /. list;
Format[circlefit[list_], StandardForm] :=
HoldForm[circlefit]["<" <> ToString@Length@list <> ">"]
This assumes that underlying data is a circle and aim is to find center and radius. There are doubtless much better ways.
For fun (and done quickly, so apologies for some inefficiencies and ugliness):
Manipulate[
pt1 = Table[{x, cy + Sqrt[rad^2 - (x - cx)^2]}, {x, -4, 4, 0.1}];
pt2 = Table[{x, cy - Sqrt[rad^2 - (x - cx)^2]}, {x, -4, 4, 0.1}];
pt = Cases[Join[pt1, pt2], {_Real, _Real}];
pts = pt + RandomReal[{-p, p}, {Length[pt], 2}];
With[{fit = circfit[pts]},
Column[{Show[
ContourPlot[Evaluate@fit["expression"], {x, -1, 8}, {y, -1, 8},
Contours -> {0}, ContourShading -> None,
ContourStyle -> Directive[Red, Thick]], ListPlot[pts],
AspectRatio -> Automatic, PlotLabel -> fit["expression"] == 0,
ImageSize -> 400, PlotRange -> {{-1, 8}, {-1, 8}}],
Grid[{{"variable", "value", "model"}, {"center", {cx, cy},
fit["center"]}, {"radius", rad, fit["radius"]}}]
}]], {p, 0.05, 1},
{cx, 0, 2},
{cy, 0, 2},
{rad, 1, 5}
]

y
in either your data or in the actual code that you used, but did not show here. Please inspect your data, and/or quit the kernel and try again. $\endgroup$