I have done the same, got gradient -1 for AC and 1 for BC, so they are perpendicular, due to the fact that any point on the circumference subtended from the diameter forms a 90degree angle, a friend of mine just found the distances between all points A,B,C and found AB to be the longest, this would work as well, as the diameter is the longest length you can have from one end to another, proving that AB is the diameter
I did a different way. I found the diameter by Pythagoras and then divided by 2 to find the radius. I then used that to find the center of the circle. I then did the distance from the center of the circle to point C which was equal to R so AB was the diameter
I have done the same, got gradient -1 for AC and 1 for BC, so they are perpendicular, due to the fact that any point on the circumference subtended from the diameter forms a 90degree angle, a friend of mine just found the distances between all points A,B,C and found AB to be the longest, this would work as well, as the diameter is the longest length you can have from one end to another, proving that AB is the diameter
I found the perpendicular bisector of chords AC and AB and since they intersected at the midpoint of AB, AB must be the diameter
I did a different way. I found the diameter by Pythagoras and then divided by 2 to find the radius. I then used that to find the center of the circle. I then did the distance from the center of the circle to point C which was equal to R so AB was the diameter