数学题:求经过p(-2,4),q(3,-1)两点,并且在x轴上截得弦长等于6的圆的方程急~

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 08:48:40
数学题:求经过p(-2,4),q(3,-1)两点,并且在x轴上截得弦长等于6的圆的方程急~
xSN@.0C;ӱeon`cFVF 1hbLA鋿NgZ(ѥ.:8δK4Z]o"*ihW5&G &w Nc0 >G gO>o?^꯲7EjZ*Ema[UQLr J`0pRԩi`JtI&sagYr TE&,(2RQ e\ :d;p>*YQ|bp\ gD- 9*C)׳ʅ9jp0m9`RvqqHs,f'fJpn7E7EX|`͌e}ީ.lvc

数学题:求经过p(-2,4),q(3,-1)两点,并且在x轴上截得弦长等于6的圆的方程急~
数学题:求经过p(-2,4),q(3,-1)两点,并且在x轴上截得弦长等于6的圆的方程
急~

数学题:求经过p(-2,4),q(3,-1)两点,并且在x轴上截得弦长等于6的圆的方程急~
设所求圆的方程为,(x-a)^2 + (y-b)^2 = c.
则,
y = 0,
(x-a)^2 = c - b^2,
x1 = a + (c-b^2)^(1/2),
x2 = a - (c-b^2)^(1/2).
2(c-b^2)^(1/2) = 6,
c - b^2 = 9,
c = b^2 + 9.
(-2-a)^2 + (4-b)^2 = c,
(3-a)^2 + (-1-b)^2 = c.
4 + 4a + a^2 + 16 - 8b + b^2 = 9 - 6a + a^2 + 1 + 2b + b^2,
10a - 10b + 10 = 0,
a = b - 1.
[3-(b-1)]^2 + (-1-b)^2 = b^2 + 9,
(4-b)^2 + (1+b)^2 = b^2 + 9,
16 - 8b + b^2 + 1 + 2b + b^2 = b^2 + 9,
b^2 - 6b + 8 = 0,
(b-2)(b-4) = 0,
b = 2,或者,b = 4.
a = 1,或者,a = 3.
c = 13,或者,c = 25.
所求圆的方程为,(x-1)^2 + (y-2)^2 = 13,或者,(x-3)^2 + (y-4)^2 = 25.