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

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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
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.

（1）设直线t与圆C1切于M（1，-3），

由圆C1：（x-4）²+（y+5)²=13，

∴圆心C1（4，-5）

由M，C1确定方程：

-3=a+b

-5=4a+b，

a=-2/3，b=-7/3

即LMC1：y=-2x/3-7/3

由点斜式，Lt：y+3=（3/2）（x-1)

y=3x/2-9/2.

（2）过C2作直线t垂线交直线t于N，

由点到直线距离就是C2的半径，

由3x-2y-9=0，及C2（3，-3）

d=C2N=|3×3+2×3-9|/√（3²+2²;）

=6/√13

∴C2:（x-3）²+（y+3)²=36/13

全部展开

(12) x2+y2-8x+10y+28=0 x^2-8x+16+y^2+10y+25=13 (x-4)^2+*(y+5)^2=13
圆心为（4,5）因为直线t与标准方程为x2+y2-8x+10y+28=0的圆C1相切在点（1，-3）
所以直线t的方程为 y-(-3)=(1-4)(x-1)/(-3+5) 2y+6=-3x+3 y=-3x/2-3/2
(13) （3，-3）到直线y=-3x/2-3/2 的距离为 /3*3-2*3+3/根号（3^2+2^2)=6/根号13
(x-3)^2+(y+3)^2=（6/根号13)^2 x^26x+9+y^2+6y+9-36/13=0
圆C2的标准方程 x^2+y^2-6x+6y+102/13=0

收起

X方-8X+16+Y方+10Y+25=41-28
（X-4）方+（Y+5）方=13
则C1（4，-5）
则直线T的斜率=-（4-1）/（-5+3）=3/2
则直线T：Y=3X/2-9/2，3X-2Y-9=0

（3，-3）到直线的距离=半径=|9+6-9|/根号（3方+2方）=6/根号13
则圆C2：（X-3）方+（Y+3）方=36/13

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