The slope is simply the tangent, so —
slope_angle = atand(p(1)) % Slope Angle (°)
Here, that comes out to be about 14.6°. The slope appears negative in the image plot because of the axes orientation being different in the images. Plotting it with both axes in their normal orientation (added at the end) shows it to be positive.
%find the circles and get the coordinates
I = imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/760726/testpic1.png');
imshow(I)
[X,Y]=meshgrid(1:size(I));
Rmin = 10;
Rmax = 20;
[cd, rd] = imfindcircles(I,[Rmin Rmax], 'ObjectPolarity', 'dark', 'Sensitivity', 0.94, 'Method','TwoStage');
viscircles(cd, rd, 'Color', 'b');
subplot(1,3,1)
imshow(I)
hold on
scatter(cd(:,1), cd(:,2),'r*')
cd = cd(cd(:,2) <315, :)
subplot(1,3,2)
imshow(I)
hold on
scatter(cd(:,1), cd(:,2),'r*')
%plot line of best fit
p = polyfit(cd(:,1),cd(:,2),1)
Bfit = polyval(p,cd(:,1));
subplot(1,3,3)
imshow(I)
hold on
scatter(cd(:,1),cd(:,2))
hold on
plot(cd(:,1),Bfit,'-r')
slope_angle = atand(p(1)) % Slope Angle (°)
figure
plot(cd(:,1),cd(:,2),'.')
hold on
plot(cd(:,1),Bfit,'-r')
hold off
axis('equal')
title(sprintf('Slope = %.1f°',slope_angle))
.
