Abstract Algebra Theory and Applications - Computer Science ...

19.3 GEOMETRIC CONSTRUCTIONS 3391. If a line contains two points in F , then it has the equation ax+by+c =0, where a, b, and c are in F .2. If a circle has a center at a point with coordinates in F and a radiusthat is also in F , then it has the equation x 2 + y 2 + dx + ey + f = 0,where d, e, and f are in F .Proof. Let (x 1 , y 1 ) and (x 2 , y 2 ) be points on a line whose coordinates arein F . If x 1 = x 2 , then the equation of the line through the two points isx−x 1 = 0, which has the form ax+by +c = 0. If x 1 ≠ x 2 , then the equationof the line through the two points is given by( )y2 − y 1y − y 1 =(x − x 1 ),x 2 − x 1which can also be put into the proper form.To prove the second part of the lemma, suppose that (x 1 , y 1 ) is the centerof a circle of radius r. Then the circle has the equation(x − x 1 ) 2 + (y − y 1 ) 2 − r 2 = 0.This equation can easily be put into the appropriate form.Starting with a field of constructible numbers F , we have three possibleways of constructing additional points in R with a compass and straightedge.1. To find possible new points in R, we can take the intersection of twolines, each of which passes through two known points with coordinatesin F .2. The intersection of a line that passes through two points that havecoordinates in F and a circle whose center has coordinates in F withradius of a length in F will give new points in R.3. We can obtain new points in R by intersecting two circles whose centershave coordinates in F and whose radii are of lengths in F .The first case gives no new points in R, since the solution of two equationsof the form ax + by + c = 0 having coefficients in F will always be in F . Thethird case can be reduced to the second case. Letx 2 + y 2 + d 1 x + e 1 x + f 1 = 0x 2 + y 2 + d 2 x + e 2 x + f 2 = 0□