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(1)∵O(0,0),A(6,2
![](https://img.yulucn.com/upload/a/ef/aef8a8a6f6084f92698e7c30afe8818d_thumb.jpg)
),
∴直线OA的方程斜率为
![](https://img.yulucn.com/upload/8/14/814468827752f1a8448c5409390a05af_thumb.jpg)
=
![](https://img.yulucn.com/upload/f/fa/ffad5e7f34e8909e875ebbc062bf194f_thumb.jpg)
,
∴线段OA垂直平分线的斜率为﹣
![](https://img.yulucn.com/upload/d/9d/d9d57ed7e0abb5f434d1a377752c4ba1_thumb.jpg)
,
又线段AO的中点坐标为(3,
![](https://img.yulucn.com/upload/1/4a/14a392c7717786e9bb04559064f577e6_thumb.jpg)
),
∴线段OA垂直平分线的方程为y﹣
![](https://img.yulucn.com/upload/d/93/d933bf17588ca6cf4627d418526df967_thumb.jpg)
=﹣
![](https://img.yulucn.com/upload/d/e6/de6988fd71cf2f0d14eb8492e36bbef5_thumb.jpg)
(x﹣3),
即
![](https://img.yulucn.com/upload/2/18/218b01b125766f9c1b275f72b04a6251_thumb.jpg)
x+y﹣4
![](https://img.yulucn.com/upload/a/d3/ad38bf86e7428639248fa587842be8c5_thumb.jpg)
=0①,
又线段OB的垂直平分线为x=4②,
∴将②代入①解得:y=0,
∴圆心C的坐标为(4,0),
又|OC|=4,即圆C的半径为4,
则圆C的方程为:(x﹣4)
2 +y
2 =16;
(2)显然切线方程的斜率存在,设切线l的斜率为k,
又切线过(2,6),
∴切线l的方程为y﹣6=k(x﹣2),即kx﹣y+6﹣2k=0,
∴圆心到切线的距离d=r,即
![](https://img.yulucn.com/upload/a/a3/aa33ac7ddc2f6f8288b8763b2e184615_thumb.jpg)
=4,
解得:k=
![](https://img.yulucn.com/upload/c/e9/ce9563eb860c8b28d3c8ccc6b208bf7e_thumb.jpg)
,
则切线l的方程为:y﹣6=
![](https://img.yulucn.com/upload/9/b6/9b6372a7435b51d3b1530dd88ee498f7_thumb.jpg)
(x﹣2);
(3)当直线l的斜率不存在时,显然直线x=2满足题意;
当直线l的斜率存在时,设斜率为k,又直线l过(2,6),
∴切线l的方程为y﹣6=k(x﹣2),即kx﹣y+6﹣2k=0,
又弦长为4
![](https://img.yulucn.com/upload/5/42/54299e088a43e60d4974f6f5a8e22c42_thumb.jpg)
,半径r=4,
∴圆心到切线的距离d=
![](https://img.yulucn.com/upload/e/9b/e9b174d230ee556a98fa81c8e51bb189_thumb.jpg)
=2,即
![](https://img.yulucn.com/upload/d/95/d95179fd6c567fa4b38f2460fdd2a64c_thumb.jpg)
=2,
解得:k=﹣
![](https://img.yulucn.com/upload/8/ee/8ee51d719f972603a9ebe2fc133228ff_thumb.jpg)
,
∴直线l的方程为y﹣6=﹣
![](https://img.yulucn.com/upload/c/ac/cac08f27491685dc8b13affcb26bfe08_thumb.jpg)
(x﹣2),即4x+3y﹣26=0,
综上,直线l的方程为x=2或4x+3y﹣26=0.
1年前
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