Triangle and Circumscribed CircleDate: 03/23/98 at 03:27:49 From: Rick Richardson Subject: Circle around a triangle I am looking for a formula that will give me the diameter/radius of a circle circumscribed around any triangle given the three outside points of the triangle (x1,y1), (x2,y2), (x3,y3): . x2,y2 . x1,y1 . x3,y3 It would be like the incircle formula except I want to find a formula for the outcircle of the triangle. Thanks very much for any help. Rick Date: 03/28/98 at 12:36:08 From: Doctor Barrus Subject: Re: Circle around a triangle Hi, Rick. I'm not familiar with the incircle formula you mentioned, but here's what I've come up with: To solve this problem I decided to find the center of the circumscribed circle and then to measure the distance from the center to one of the points. This would give the circle's radius. Geometrically, to find the center of a circle that passes through three points, you draw two lines, each connecting two of the points. You can construct perpendicular bisectors to these lines. The center of the circle will be where the perpendicular bisectors meet. This happens because if you connect two points A and B with a line segment AB and construct the perpendicular bisector to AB (let's call it d), every point on d is equidistant from A and B. Now let's call the 3 points of our triangle A, B, and C. We'll connect points A and B to form segment AB, and we'll connect B and C to form segment BC. The perpendicular bisectors of AB and BC (we'll call them d and f, respectively), should meet at some point O (as long as A, B, and C don't all lie in a straight line). Since all points on d are equidistant from A and B, and all points on f are equidistant from B and C, point O is equidistant from all three, so a circle with center O passing through one of the points will also pass through the other two. Algebraically, this problem's a mess. I made all my calculations on the symbolic math software Maple V (if I had tried to do it by hand, I'd still be working on it ;) ). First of all, I found the center point of the circle that circumscribes the triangle formed by the three points (the outcircle). I found the slopes of the line that passes through (x1,y1) and (x2,y2) and the line that passes through (x2,y2) and (x3,y3). I did this by using the general formula: slope = (change in y)/(change in x). Then I found the slopes of the lines that are perpendicular bisectors to these lines by noting that if two lines with slopes m1 and m2 are perpendicular, then m1*m2 = -1 and m2 = -1/m1 Then I used the general linear formula y = mx+b, plugged in the slopes for m and the coordinates of the midpoints of the 2 segments joining points 1 and 2, and 2 and 3, for x and y. [For two points (f,g) and (h,j) the midpoint coordinates are ((f+h)/2, (g+j)/2).] I then solved for the b's. Needless to say, it was a very messy calculation. Once I found the equations of the perpendicular bisector lines, I could find the coordinates of the outcircle's center by taking my two equations eq. 1 (bisector of (x1,y1)(x2,y2)) y = m1*x + b1 eq. 2 (bisector of (x2,y2)(x3,y3)) y = m2*x + b2 and setting the two y-coordinates equal to each other (solving this gives the point on both lines with the same x and y-coordinates, in other words, the point that both lines share). So I had m1*x + b1 = m2*x + b2 I solved for x (or rather, had the computer solve for x, since m1, m2, b1, and b2 were huge expressions), the x-coordinate of the lines' intersection. I substituted my answer into equation 1 (either equation 1 or equation 2 will work) to find out the y-coordinate of the intersection. Then I used the distance formula: Distance between (a,b) and (c,d) = sqrt((a-c)^2 + (b-d)^2) to find the distance between the center point and (x1, y1). (I could have used any one of the three original points.) This, finally, was the radius. After simplifying the nasty expression Maple gave me, I arrived at the following: 1 D12 * D23 * D31 radius of the outcircle = - * ---------------------------------- 2 y1(x3-x2) + y2(x1-x3) + y3(x2-x1) Where D12 is the distance between (x1,y1) and (x2,y2), D23 " " " " (x2,y2) " (x3,y3), and D31 " " " " (x3,y3) " (x1,y1), which we can find by using the distance formula above. This formula seems to work in all the cases I've tried. If there's a certain form you're looking for, you can probably manipulate the above formula to match that form. I hope this has helped. Good luck! -Doctor Barrus, The Math Forum Check out our Web site http://mathforum.org/dr.math/ |
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