Question

圆满足截Y轴所得弦长为2 被X轴分成两段圆弧 弧长比3：1 圆心到直线L：X-2Y=0距离为五分之根号五 求该园方

Answer 1

设圆心C(a, b), 圆方程: (x-a)^2 + (y - b)^2 = r^2 (r > 0)
C到直线L：X-2Y=0距离: |a - 2b|/√(1+2^2) = |a - 2b|/√5 = √5/5
|a - 2b| = 1
a - 2b = 1或a - 2b = -1

x = 0: (x-a)^2 + (y - b)^2 = r^2 => (y-b)^2 = r^2 - a^2
y = b±√(r^2 - a^2)
圆截Y轴所得弦长为[b + √(r^2 - a^2)] - [b - √(r^2 - a^2)] = 2√(r^2 - a^2) =2
r^2 - a^2 = 1 (1)

设圆与X轴的交点为A, B, 弧长比3：1, 角ACB = 90˚
y = 0: (x-a)^2 + (y - b)^2 = r^2 => (x-a)^2 = r^2 - b^2
x = a ±√(r^2 - b^2)
|AB| = [a + √(r^2 - b^2)] - [a - √(r^2 - b^2)] = 2√(r^2 - b^2)
|AB|^2 = |AC|^2 + |BC|^2 = 2r^2 = 4(r^2 - b^2)
r^2 = 2b^2 (2)

(1) a - 2b = 1 (3)
由(1)(2)(3), a = -1, b = -1, r = √2
(x+1)^2 + (y+1)^2 = 2

(2) a - 2b = -1 (4)
由(1)(2)(4), a = 1, b = 1, r=√2
(x-1)^2 + (y-1)^2 = 2