Question

已知圆C1:x2+y2-2x-4y-13=0,C2:x2+y2-2ax-6y+a2+1=0,(a>0)相外切,且直线L:

Answer 1

(1)C1 :x2+y2-2x-4y-13=0==>(x-1)²+(y-2)²=18

C2:x2+y2-2ax-6y+a2+1=0 ==> (x-a)²+(y-3)²=8

相外切==> |C1C2|=r1+r2

√[(a-1)²+1]=3√2+2√2=5√2

==>a²-2a-48=0 ==>a1=8,a2=-6

∵a>0,∴a=8

圆C2的标准方程 (x-8)²+(y-3)²=8

(2)直线L:mx+y-7=0与C2相切

圆心到直线的距离等于圆的半径：d=|8m+3-7|/√(m²+1)=2√2

==>7m²-8m+1=0

==>m=1,或m=1/7