Question

在平面直角坐标系xOy中，曲线y＝x^2-6x+1与坐标轴的交点都在圆C上。（1）求圆C的方程

2.若圆C与直线x-y+a=0交于A、B两点，且OA垂直OB求a的值
求解释第2问

Answer 1

(1)
曲线y=x² -6x+1与y轴的交点: D(0, 1)
y = x² -6x+1 = 0, x = 3±2√2, 与x轴的交点: A(3-2√2, 0), B(3+2√2, 0)
曲线y=x² -6x+1为抛物线，对称轴为x = 3
显然圆心C在对称轴上，设C(3, b)
CA = CD = r
CA² = CD²
(3 - 3 + 2√2)² + (b - 0)² = (3- 0)² + (b - 1)²
8 + b² = 9 + b² - 2b + 1
b = 1
C(3, 1)
C, D的纵坐标相同，距离为横坐标之差, r = 3-0 = 3
圆C的方程: (x - 3)² + (y - 1)² = 9

(2)OAB为直角三角形, AB = √(OA² + OB²) = √2r = 3√2
三角形OAB的面积S= OA*OB/2 = AB*AB上的高h/2= 3*3/2 = 3√2*h/2
h = 3/√2
h为C与直线x-y+a=0距离 = |3 - 1 + a|/√2 = |a+2|/√2= 3/√2
|a+2|= 3
a= 1或a = -5

Answer 2

见http://zhidao.baidu.com/question/482398299.html?oldq=1

Answer 3

a=-1