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
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.
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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