原题是英文In questions 12-13,line t is tangent to the circle C1,o

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原题是英文
In questions 12-13,line t is tangent to the circle C1,of equation x2+y2-8x+10y+28=0 at (1,-3).t is also tangent to the circle C2 whose center is (3,-3).
12.Find an equation of t.
13.Find the standatd equation of C2.
翻译：
在12,13题中,直线t与标准方程为x2+y2-8x+10y+28=0的圆C1相切在点（1,-3）.同时直线t也和圆心为（3,-3）的圆C2相切.
12.写出直线t的方程
13.写出圆C2的标准方程

直线t与标准方程为x2+y2-8x+10y+28=0的圆C1相切在点（1,-3）.同时直线t也和圆心为（3,-3）的圆C2相切.

12. 写出直线t的方程

13. 写出圆C2的标准方程 12.由x2+y2-8x+10y+28=0配方得（x-4)²+（y+5)²=13∴圆C1的圆心为D（4,-5）、半径是√13.由题设：直线t与标准方程为x2+y2-8x+10y+28=0的圆C1相切在点C（1,-3）.由斜率公式得直线CD的斜率为：-2/3,∴直线t的斜率为：3/2,由点斜式得直线t的方程为：3x-2y-9=0. 13.题设：直线t也和圆心为（3,-3）的圆C2相切.
由点到直线的距离公式得点为（3,-3）到直线 t：3x-2y-9=0的距离是：d=|3·3+2·3-9|/√(3²+（-2）²)=6√13/13.得圆C2的半径r=6√13/13.∴圆C2的标准方程为：（x-3)²+（y+3)²=36/13.如图.

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