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    <p>I have to say, if you're following the link you listed, it would've helped to keep the variable names the same. We could understand the algorithm much better seeing x1, y1, x2, y2, x3, y3 instead of p1, p2, p3, p4, alpha and beta. In fact, I don't see much in your algorithm that matches the link. I'm not trying to be as harsh as the comments were (and if you're worried about switching float to double, that was a perfectly good case for a typedef), but debugging algorithms is easiest when you don't have to convert variable names.</p> <p>I would recommend simply using what they give you for h and k in the link, which is namely done by calculating determinants of 3x3 matrices. You can find lots of <a href="http://www.wikihow.com/Find-the-Determinant-of-a-3X3-Matrix" rel="nofollow">references</a> for that.</p> <p>I'd make two functions, as follows:</p> <pre><code>float calculateH(float x1, float y1, float x2, float y2, float x3, float y3) { float numerator = (x2*x2+y2*y2)*y3 - (x3*x3+y3*y3)*y2 - ((x1*x1+y1*y1)*y3 - (x3*x3+y3*y3)*y1) + (x1*x1+y1*y1)*y2 - (x2*x2+y2*y2)*y1; float denominator = (x2*y3-x3*y2) - (x1*y3-x3*y1) + (x1*y2-x2*y1); denominator *= 2; return numerator / denominator; } float calculateK(float x1, float y1, float x2, float y2, float x3, float y3) { float numerator = x2*(x3*x3+y3*y3) - x3*(x2*x2+y2*y2) - (x1*(x3*x3+y3*y3) - x3*(x1*x1+y1*y1)) + x1*(x2*x2+y2*y2) - x2*(x1*x1+y1*y1); float denominator = (x2*y3-x3*y2) - (x1*y3-x3*y1) + (x1*y2-x2*y1); denominator *= 2; return numerator / denominator; } </code></pre> <p>Then your is_formCircle would simply be:</p> <pre><code>float is_formCircle(float x1, float y1, float x2, float y2, float x3, float y3) { float h = calculateH(x1, y1, x2, y2, x3, y3); float k = calculateK(x1, y1, x2, y2, x3, y3); printf("x=%f y=%f\n",h,k); } </code></pre> <p>There are tons of ways to optimize this, and there's a chance I typoed any of the determinant calculations, but it should get you going.</p>
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    1. POfind center of circle when three points are given
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    1. USScott Mermelstein
    1. USScott Mermelstein
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    1. COThank You.. I did same thing using paper and pen and formulize the thing in my code but not working but this is working perfectly..
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      1. USavinashse
    2. COYou're welcome. I notice that your code is actually a boolean check, to see if you can form a circle from the three points. The answer (as you can guess from Dr_Sam's explanation) is always "yes" - unless the three points are colinear. If they are, in this calculation, the determinant in the denominator will evaluate to 0.
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      1. USScott Mermelstein
    3. COYes I thought about this, three points will always form circumcircle unless they are collinear, what I am doing is I have to check forth point is lying inside,outside or on the circle, For e.g, let there be for point A,B,C,D . I am finding the center of circle A,B,C and then calculating the radius(R) and then calculating |AD|. If (|AD| > R) point D is outside the circle, so Is there any fast way to do it..?
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