Question

一个圆与直线x+3y-26=0切于点（8,6）,且经过点（-2,-4）,求圆的方程

Answer 1

C(a,b)

L:x+3y-26=0,k(L)=-1/3

k(CB)=-1/k(L)

(b-6)/(a-8)=3

3a-b-18=0.(1)

AB的中点M(3,1)

k(CM)*k(AB)=-1

[(b-1)/(a-3)]*[(6+4)/(8+2)]=-1

a+b-4=0.(2)

(1),(2):

a=5.5,b=-1.5

C(5.5,-1.5)

r^2=62.5

(x-5.5)^2+(y+1.5)^2=62.5