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

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| = 1a - 2b = 1或a - 2b = -1x = 0:(x-a)^2 + (y - b)^2 = r^2 => (y-b)^2 = r^2 - a^2y = 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) =2r^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^2x = 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