Radius and coordinates of the center for a big circle in a bidimensional image

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I have a bidimensional (N x N) matrix of pixels like the one shown in the attached figure. Black areas contain zero values, while other areas contain random non-zero values. How could I determine the radius (in pixels) and coordinates of the center for the big circle? Consider that its center doesn't correspond to the center of the matrix, like in the figure.

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Image Analyst
Image Analyst 2016-2-11
编辑:Image Analyst 2016-2-11
Get a binary image by thresholding. Then fill holes, label, and call regionprops().
binaryImage = grayImage > 10; % or whatever value works.
% Fill holes
binaryImage = imfill(binaryImage, 'holes');
% Label
labeledImage = bwlabel(binaryImage);
% Measure
measurements = regionprops(labeledImage, 'EquivDiameter');
% Extract
theRadius = measurements.EquivDiameter / 2;
By the way, instead of the very verbose and overly complicated "bidimensional (N x N) matrix of pixels", most people would simply just say "2D image".
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matnewbie
matnewbie 2016-2-11
Thanks for your answer. I modified the original post, since I forgot to say that I am also interested in determining the coordinates of the center for the big circle. Is it possible to obtain the same results without using functions from the Image Processing Toolbox? Since I have a matrix of values, is it possible to find the coordinates of the center by minimizing the number of black pixels (zero values) from the edges of the black square?

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Anand
Anand 2016-2-11
You can use the awesome imfindcircles function to do this as well.
Here's some code that should help:
% Read the image
I = imread('http://www.mathworks.com/matlabcentral/answers/uploaded_files/45140/figformatlab.png');
% Find circles in the image
radiusRange = [20 30]; % You need an approximate range of what you expect the radius to be.
[centers,radii] = imfindcircles(I, radiusRange, 'ObjectPolarity', 'dark','Sensitivity',.9);
% Display found circles
imshow(I)
viscircles(centers,radii);
The variables centers and radii hold what you are looking for.

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