circle geometry

👨🏻‍🏫 學術台數學

[匿名]

2013-03-28 12:29 am

1. P and Q are the end-points of a fixed length chord moving along the circumference of a fixed circle x^2+y^2+4x-6y+3=0.If PQ=4,find the equation of the locus of the mid-point M of PQ.

2.Consider the circle S: x^2+y^2+2x-2y-3=0 with centre C.Two tangents are
drawn to the circle S from the point A(2,0) outside the circle.

a.Find the equations of the two tangents.

b.Let P and Q be the points of the contact of the two tangents with positive slope
and negative slope respectively.

i.Find the coordinates of P and Q.

ii.What kind of quadraulateral is PCQA?Explain your answer.

唔該詳細解釋及列出清晰既步驟!!

更新1:

我想問l 1+m+2m l/開方(1+m^2)是一條咩式，我都未學過，可唔可以用其他方法計?唔該哂!!!

回答 (2)

用戶9112

2013-03-28 5:18 am

✔ 最佳答案

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圖片參考：http://imgcld.yimg.com/8/n/HA05107138/o/20130327211633.jpg

圖片參考：http://imgcld.yimg.com/8/n/HA05107138/o/20130327211752.jpg

2013-03-28 18:49:28 補充：
The distance of a point (h,k) from a line ax+by+c=0 is
|ah+bk+c|/sqrt(a^2+ b^2)
This is a common equation. I am surprised that you have not learnt it.

2013-03-28 18:51:29 補充：
Otherwise sub the equation of the tangent to the equation of the circle:
x^2+y^2+2x-2y-3=0
x^2+ (mx-2m)^2+2x-2(mx-2m)-3=0
x^2+m^2 x^2-4m^2 x+4m^2+2x-2mx+4m-3=0
(1+m^2 ) x^2+(2-2m-4m^2 )+(4m^2+4m-3)=0

2013-03-28 18:51:40 補充：
For tangency, discriminant=0, so
(2-2m-4m^2 )^2-4(1+m^2 )(4m^2+4m-3)=0
(2m^2+m-1)^2-(1+m^2 )(4m^2+4m-3)=0
4m^4+m^2+1+4m^3-4m^2-2m-4m^2-4m+3-4m^4-4m^3+3m^2=0
-4m^2-6m+4=0
-2(2m-1)(m+2)=0
So m=1/2 or m=-2

👍 0 👎 0

[匿名]

2013-03-28 6:45 am

我想問l 1+m+2m l/開方(1+m^2)是一條咩式，我都未學過，可唔可以用其他方法計?唔該哂!!!

👍 0 👎 0

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