find values for the equation of a circle

I have the equation of a circle:

(x - a)^2 + (y - b)^2 - r^2 = 0

a, b and r are known values but I would like to find the values of x and y from a predetermined coordinate matrix which satisfies the equation.

example: a = 2, b = 2, r = 2

A = 

    1 2
    1 3
    2 4

hence the last row in A(2,4) would satisfy the equation. i.e. x = 2, y = 4

this is what I've tried so far:

a = 2;
b = 2;
r = 2;
x = find(A(:,1);
y = find(A(:,2);
solve((x - a)^2 + (y - b)^2 - r^2 == 0)

Answer 1

[Roger Stafford]

 a = 2; b = 2; r = 2;
 A = [1 2;1 3;2 4];
 B = A((A(:,1)-a).^2+(A(:,2)-b).^2==r^2,:);

B contains the coordinate pairs, if any, that lie on the circle. Note that if you need this for values of a, b, and c which are not integers, it is best to allow a small tolerance for round-off errors rather than demanding exact equality in the equation.

[Duncan]

hi, thanks for the reply. what do you mean by coordinate pairs? I ran the code and it only shows up with a single value 2

[Roger Stafford]

That was an error. I have now corrected it. My apologies.

[Duncan]

I have another question related to this. Say:

A = 1 2
    1 3
    2 4
    4 2

then the last two values would satisfy and hence display 2,4 and 4,2.

What I would like to know is how this code would change if i set limits. Say I want to display only the value that is within the limits of +/-1 of a & b.

i.e. it would only display 2,4