两圆x²+y²=m和x²+y²+6x-8y-11=0有公共点,则实数m的范围是A.m121C.1≤m≤121D.1
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![两圆x²+y²=m和x²+y²+6x-8y-11=0有公共点,则实数m的范围是A.m121C.1≤m≤121D.1](/uploads/image/z/3800845-37-5.jpg?t=%E4%B8%A4%E5%9C%86x%26%23178%3B%2By%26%23178%3B%3Dm%E5%92%8Cx%26%23178%3B%2By%26%23178%3B%2B6x-8y-11%3D0%E6%9C%89%E5%85%AC%E5%85%B1%E7%82%B9%2C%E5%88%99%E5%AE%9E%E6%95%B0m%E7%9A%84%E8%8C%83%E5%9B%B4%E6%98%AFA.m121C.1%E2%89%A4m%E2%89%A4121D.1)
两圆x²+y²=m和x²+y²+6x-8y-11=0有公共点,则实数m的范围是A.m121C.1≤m≤121D.1 两圆x²+y²=m和x²+y²+6x-8y-11=0有公共点,则实数m的范围是A.m121C.1≤m≤121D.1 C,利用圆心距和半径的关系知,6-5 <= 根号m <= 6+5 c
两圆x²+y²=m和x²+y²+6x-8y-11=0有公共点,则实数m的范围是
A.m<1
B.m>121
C.1≤m≤121
D.1
方程 x^2+y^2+6x-8y-11=0 配方得 (x+3)^2+(y-4)^2=36,因此圆心(-3,4),半径为 6 ,
圆心距 d=√(9+16)=5,
因为两圆有公共点,因此 |r1-r2| ≤ d ≤ r1+r2 ,
即 |√m-6| ≤ 5 ≤ √m+6 ,
解得 1 ≤ m ≤ 121 .
选 C