R,r是两圆的半径(R>r),d是两圆的圆心距,若方程x²-2Rx+r²=d(2r-d）有等根...

题目

R,r是两圆的半径(R>r),d是两圆的圆心距,若方程x²-2Rx+r²=d(2r-d）有等根...
则以R、r为半径的两圆的位置关系是

答案

x² - 2Rx + r² = d(2r-d)x² - 2Rx + r² - 2dr + d² = 0x² - 2Rx + R² + r² - 2dr + d² - R² = 0(x - R)² + (r² - 2dr + d² - R²) = 0 x = R...