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Inquestions12-13,linetistangenttothecircleC1,ofequationx2+y2-8x+10y+28=0at(1,-3).tisalsotangent

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

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

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.

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

(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 ...

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(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