Question

已知圆x2+y2=m与圆x2+y2+6x-8y-11=0相内切，则实数m的值为______．

Answer 1

已知圆⊙1x²+y²=m与圆⊙2x²+y²+6x-8y-11=0相内切

x²+y²+6x-8y-11=(x+3)^2+(y-4)^2=36

⊙1圆心(0,0)半径r1=√m ,⊙2圆心(-3,4),半径r2=6

∴圆心距=√(-3)^2+4^2 =5

∴|√m - 6|=5

①√m=11 ,m=121

②√m=1 ,m=1

祝学习进步！O(∩_∩)O~