圆经过点a(0,2),b(3,-1),c(4,0);(2)圆心在c(3,-5),圆与直线x-7y+2=0相切.
问题描述:
圆经过点a(0,2),b(3,-1),c(4,0);(2)圆心在c(3,-5),圆与直线x-7y+2=0相切.
答
圆(x-a)^2+(y-b)^2=r^2经过点A(0,2),B(3,-1),C(4,0)所以a^2+(2-b)^2=r^2 (1)(3-a)^2+(-1-b)^2=r^2 (2)(4-a)^2+b^2=r^2 (3)由(1)(2)得a=b+1 (4)由(1)(3)得b=2a-3 (5)由(5)代入(4)...