ce a.maths CO-GEOM(CIRCLE)

Andrew

2007-12-04 4:40 am

1. Find the common chord of the circles x^2+y^2-4y-3=0 and x^2+y^2-4x-2y-5=0.
2.Find the equation of the circle which passes through the points of intersection of the line x-2y=0 and the circle x^2+y^2+4x-2y+1=0 and whose centre lies on the x-axis.
3.Prove that the circles x^2+y^2+2x-8y+8=0 and x^2+^2-6x-2y+6=0 touch each other, and find their common tangent at the point of contact.

回答 (1)

Gabriella Montez

2007-12-04 5:42 am

✔ 最佳答案

1) C1: x^2+y^2-4y-3=0

C2: x^2+y^2-4x-2y-5=0

Common chord: ( C1 - C2 )

x^2 + y^2 - 4y - 3 - x^2 - y^2 + 4x + 2y + 5 = 0

2x - y + 1 = 0

2) Family of circle:

x^2 + y^2 + 4x - 2y - 1 + k ( x - 2y ) = 0 where k is real

Rearranging,

x^2 + y^2 + ( k + 4 )x + ( - 2 - 2k )y - 1 = 0

The centre: [ -(k+4)/2, (k+1)]

So,

k + 1 = 0

k = -1

Then,

x^2 + y^2 + ( -1 + 4 )x + ( - 2 + 2 )y - 1 = 0

x^2 + y^2 + 3x - 1 = 0

3) The centre of x^2+y^2+2x-8y+8=0: ( -1 . 4 )

The radius: (1/2)[sqr36] = 3

The centre of x^2+^2-6x-2y+6=0: ( 3 , 1 )

The radius: (1/2)[sqr16] = 2

The distance between radius: 3 + 2 = 5

The distance between centres: sqr [ ( - 1 - 3 )^2 + ( 4 - 1 )^2 ] = 5

So the circles touch each other externally.

Common tangent:

x^2 + y^2 - 6x - 2y + 6 - x^2 - y^2 - 2x + 8y - 8 = 0

4x - 3y + 1 = 0

參考: My Maths Knowledge

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