高中数学问题：圆的方程

(x-a)^2+(y-b)^2=r^2
(a,b),r
5a-3b-8=0
/b/=/a/=r
b^2=a^2
b^2-a^2=0
(b-a)(b+a)=0
b=aorb=-a
1.b=a
5a-3a-8=0
2a=8
a=4,b=4
b=-a
r=4
5a-3(-a)-8=0
5a+3a-8=0
8a=8
a=1
b=-1
r=1
(x-4)^2+(y-4)^2=16or(x-1)^2+(y+1)^2=1
(2)(x-a)^2+(y-b)^2=r^2
a^2+(4-b)^2=r^2(1)
(4-a)^2+(B-b)^2=r^2(2)
a-2b-2=0(3)
(2)-(1) 16-8a+B^2-2Bb-16+8b=0
8a+(2B-8)b=B^2
a=2b+2
8(2b+2)+(2B-8)b=B^2
b=(B-4)/2
a=B-2
r^2=(5B^2-40B+160)/4
(x-(b-2))^2+(y-(b-4)/2)^2=(5b^2-40b+160)/4