1.設(shè)直線為y = 3x + b,3x -y + b = 0
圓心C(1,0),半徑r = 1
C與3x -y + b = 0的距離d = |3 + b|/√(3²+1²) = |3+b|/√10 = r = 1
b+3 = ±√10
b = ±√10 - 3
y = 3x ±√10 - 3
2.x + y - 1 = 0
y = 1 - x
斜率為-1
設(shè)直線為y = -x + b,x + y - b = 0
C與x + y - b = 0的距離d = |1 + 0 - b|/√(1²+1²) = |b-1|/√2 = r = 1
b - 1 = ±√2
b = 1 ±√2
y = -x + 1 ±√2