考虑过原点的两圆乘积:
圆心为$(x_1,y_1)(x_2,y_2)$,动点为$(-x_1 \cos \phi +x_1+y_1\sin \phi,-y_1\sin \phi-y_1 \cos \phi+y_1)$与$(-x_2 \cos \theta +x_2+y_2\sin \theta,-y_2\sin \theta-y_2 \cos \theta +y_2)$,两动点之积为(x,y).
令$u=\tan{\frac\phi2},v=\tan{\frac\theta2}$,在$u^2 v^2 x+u^2 x+4 u v x_1 x_2-4 u v y_1 y_2-4 u x_1 y_2-4 u x_2 y_1+v^2 x-4 v x_1 y_2-4 v x_2 y_1+x-4 x_1 x_2+4 y_1 y_2=u^2 v^2 y+u^2 y+4 u v x_1 y_2+4 u v x_2 y_1+4 u x_1 x_2-4 u y_1 y_2+v^2 y+4 v x_1 x_2-4 v y_1 y_2-4 x_1 y_2-4 x_2 y_1+y=0$中消去v得$u^2 x^2+u^2 y^2-4 u x x_1 y_2-4 u x x_2 y_1+4 u x_1 x_2 y-4 u y y_1 y_2+x^2-4 x x_1 x_2+4 x y_1 y_2-4 x_1 y y_2-4 x_2 y y_1+y^2$,令$Δ_u$≥0得$-4 x^2 \text{x1}^2 \text{y2}^2-8 x^2 \text{x1} \text{x2} \text{y1} \text{y2}-4 x^3 \text{x1} \text{x2}-4 x^2 \text{x1} y \text{y2}-4 x^2 \text{x2}^2 \text{y1}^2-4 x^2 \text{x2} y \text{y1}+2 x^2 y^2+4 x^3 \text{y1} \text{y2}+x^4+8 x \text{x1}^2 \text{x2} y \text{y2}+8 x \text{x1} \text{x2}^2 y \text{y1}-4 x \text{x1} \text{x2} y^2-8 x \text{x1} y \text{y1} \text{y2}^2-8 x \text{x2} y \text{y1}^2 \text{y2}+4 x y^2 \text{y1} \text{y2}-4 \text{x1}^2 \text{x2}^2 y^2+8 \text{x1} \text{x2} y^2 \text{y1} \text{y2}-4 \text{x1} y^3 \text{y2}-4 \text{x2} y^3 \text{y1}-4 y^2 \text{y1}^2 \text{y2}^2+y^4\le0$. |