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问题:如图1,已知圆C:x2+y2=r2与直线l:y=kx+m没有公共点,设点P为直线l上的动点,过点P作圆C的两条切线,A、B为切点。证明:直线lAB恒定过点Q。分析:利用我们常用的一个结论:若点P(x0,y0)是圆x2+y2=r2外一点,则过点P作圆的两条切线,切点分别为A、B,则过A、B两点的直线方程为:x0·x+y0·y=r2。
Problem: As shown in Figure 1, it is known that there is no common point between circle C: x2 + y2 = r2 and line l: y = kx + m. Let point P be the moving point on line l and point P as the two tangent of circle C , A, B for the cut-off point. Proof: Line lAB constant over point Q. Analysis: The use of our common conclusion: If the point P (x0, y0) is a circle x2 + y2 = r2 a point, then the point P for the two tangent of the circle, the point of cut were A, B, then A, B two-point equation of the line is: x0 · x + y0 · y = r2.