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1.6. Quadratic field extensions 21 (b) This is essentially the definition of an F -circle. If h, k, r2 ∈ F , then r2 is the distance from the origin to (r2, 0), which is an F -point. The converse is just as easy. Note that an F -line will pass through points that are not F -points. For example, the line y = x is a Q-line, but it passes through ( 2, √ 2). The reader can check that if an F -line passes through a point where one coordinate belongs to F , then the other coordinate also belongs to F . Also a circle can be an F -circle without the radius belonging to F . For example, x2 + y2 = 2 is a Q-circle but the radius is not rational. 1.6.9. Lemma. Let F be a field. (a) The point of intersection of two F -lines is an F -point. (b) The points of intersection of an F -line and an F -circle have coordinates belonging to some quadratic extension of F . (c) The points of intersection of two F -circles have coordinates belonging to some quadratic extension of F . Proof. (a) Let y = m1x + b1 and y = m2x + b2 be the equations of two F -lines. According to Lemma 1.6.8, m1,b1,m2,b2 ∈ F . To find the point of intersection of these two lines we must solve these two equations simultane- ously for x and y. The solutions are x = − b1 − b2 m1 − m2 , y = m1 − b1 − b2 m1 − m2 + b1, which belong to F . (b) Let y = mx+b and (x−h)2+(y−k)2 = r2 be the F -line and F -circle so m, b, h, k, r2 ∈ F . Again we must solve these equations simultaneously. If we simplify the equation (x − h)2 + (mx + b − k)2 − r2 = 0, we get Ax2 + Bx + C = 0, where A = 1 + m2, B = −2h + 2mb − 2km, C = h2 − 2kb + k2 − r2. Note that A, B, C ∈ F . Applying the quadratic formula to solve for x we get that x ∈ F ( √ B2 − 4AC). It follows that y = mx + b also belongs to this quadratic extension of F . (c) We have the equations (x − h1)2 + (y − k1)2 = r2 1 and (x − h2)2 + (y − k2)2 = r2 2 with h1,h2,k1,k2,r1,r2 2 2 in F , and these equations must be