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Last edited by hbghlyj 2023-1-30 00:33GroebnerBasis[{(x - x1)^2 + (y - y1)^2 - (r - r1)^2, (x - x2)^2 + (y -y2)^2 - (r - r2)^2, (x - x3)^2 + (y - y3)^2 - (r - r3)^2}, {r, x1, y1, x2, y2, x3, y3, r1, r2, r3}, {x, y}]
Factor[Discriminant[%[[1]],r]]
\[-16 \left(r_1^2-2 r_1 r_2+r_2^2-x_1^2+2 x_1 x_2-x_2^2-y_1^2+2 y_1 y_2-y_2^2\right)\]\[ \left(r_1^2-2 r_1 r_3+r_3^2-x_1^2+2 x_1 x_3-x_3^2-y_1^2+2 y_1 y_3-y_3^2\right) \]\[\left(r_2^2-2 r_2 r_3+r_3^2-x_2^2+2 x_2 x_3-x_3^2-y_2^2+2 y_2 y_3-y_3^2\right)\]\[ (-x_1 y_2+x_1 y_3+x_2 y_1-x_2 y_3-x_3 y_1+x_3 y_2)^2\]
突然想到...Apollonius的作圆与三圆相切的问题从代数上可以转换为平移等价的三个正圆锥相交吧 |
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