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过某曲线上一点P向圆A引两条切线,若这两条切线又与某曲线G(或直线)分别交于B,C两个点,求解相关问题.这是近几年高考或统考中的常见题.本文介绍一个有效的方法──同一法.下面分三种情况说明这类问题的统一解法.1.若曲线G为直线解决这类问题的方法是:先设出B,C的横(或纵)坐标,写出切线PB的方程,由直线和圆相切的条件可得点B的横(或纵)坐标所满足的一元二次方程,同理可得出点C的横(或纵)坐标所满足的
If the two tangent lines are again intersected with a certain curve G (or straight line) at points B and C respectively, the relevant problem will be solved. This is the result of the college entrance examination or the entrance examination in recent years In the common problem.This article describes an effective method ─ ─ the same method.The following three cases illustrate the unified solution of such problems.1.If the curve G is a straight line to solve such problems is: first set B, C Horizontal (or vertical) coordinates, write the equation of the tangent PB, tangent from the straight line and the circle can be horizontal (or vertical) coordinates of the point B to meet the quadratic equation quadratic, the same point can be drawn C horizontal (Or vertical) coordinates to meet