(x-6)^2+(y+3)^2=r^2和(x-2)^2+(y+1)^2=r^2和(x+1)^2+(y+10)^2=r^2怎么解?注重过程。 两圆(x+1)^2+(y-1)^2=r^2和(x-2)^2+...

(x-a)^2+(y-b)^2=r^2\u662f\u4e00\u4e2a\u5706\u3002 (x-a)^2+(y-b)^2+(z-c)^2=

\u6b63\u786e\u7ed3\u679c\u662f \u8fd9\u4e2a\u5f0f\u5b50\u8868\u8fbe\u7684\u56fe\u5f62\u4e09\u7ef4\u4e16\u754c\u65e0\u6cd5\u60f3\u8c61\uff0c\u4ee5\u6211\u4eec\u7684\u8ba4\u77e5\u4e0d\u5b58\u5728\u3002

\u56e0\u4e3az\u7684\u8868\u8fbe\u5f0f\u6ca1\u6709\u5b9e\u6570\u89e3\u3002\u800c\u4e14\u8fd9\u4e2a\u51fd\u6570\u672c\u8eab\u662f\u4e09\u6b21\u65b9\u7a0b\u3002



\u89e3:\u4e24\u76f8\u4ea4\u5706\u7684\u4ea4\u70b9\u5173\u4e8e\u8fde\u5fc3\u7ebf\u5bf9\u79f0
\u8bbeQ\u5750\u6807(a,b)
\u8fde\u5fc3\u7ebf\u65b9\u7a0b\u4e3ax+y=0
\u2234a+1+b+2=0
(b-2)/(a-1)=1
\u89e3\u5f97a=-2,b=-1
\u2234Q(-2,-1)

你应该设圆方程为x^2+y^2+dx+ey+f=0,代人三个点坐标,可以得到三元一次方程组,从而方便求解。
现在解你给定的方程组,正确的方法是:
(x-6)^2+(y+3)^2=(x-2)^2+(y+1)^2 (x+1)^2+(y+10)^2=(x-6)^2+(y+3)^2
展开化简得到2x-y-10=0 x+y+4=0
解得x=2 y=-6
最后求得r=5

圆的标准方程
(x-6)^2+(y+3)^2=r^2 圆心(6,-3) 半径r
(x-2)^2+(y+1)^2=r^2 圆心(2,-1) 半径r
(x+1)^2+(y+10)^2=r^2 圆心(-1,-10) 半径r

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