Finding center of point layer in ArcGIS Pro

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I have a point layer that has Lat/long coordinates. How to find the center of a given point layer. For more understanding, I attached a drawing.

How is it possible to do this?

I tried to use the tools "Mean center", "FeatureToPoint", but they give false results and find exactly not the center, can there be other tools or can you somehow find it using python, I have an attribute table that contains only the lat/long coordinates (-180 :180) (-90: 90). Maybe you can find it by any formulas or libraries?

I tried to do as you said, but the center also appears in the wrong place, maybe I made the wrong polygon? I'll add a photo to the question.

enter image description here

Yes, I created the CONVEX_HULL polygon. I put the dot layer in Google drive. The link in the question. https://drive.google.com/drive/folders/1tCOYZLsiBFGxCjs5DuQ7_mt42BRSJGuA?usp=sharing

enter image description here

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1 Answers

I would try to use Minimum Bounding Geometry (Data Management) with geometry type - CONVEX_HULL to create polygon feature from your input points.

arcpy.management.MinimumBoundingGeometry(in_features, out_feature_class, {geometry_type}, {group_option}, {group_field}, {mbg_fields_option})

And then you can use Feature To Point (Data Management) and create centroid from convex polygon:

arcpy.management.FeatureToPoint(in_features, out_feature_class, {point_location})

Edit:

Yes, it looks like the polygon was not created correctly. Did you create polygon using a convex hull? Because when using convex hull a polygon should be created as in the picture below. Can you add your point layer? I would try where the problem is.

enter image description here

You need to change the coordinate system of your data (using Project (Data Management)). Select the UPS North instead of the WGS 1984 you used before. After changing the coordinate system convex hull works properly - picture below.

arcpy.management.Project(in_dataset, out_dataset, out_coor_system, {transform_method}, {in_coor_system}, {preserve_shape}, {max_deviation}, {vertical})

enter image description here

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