AMATHS CIRCLE 急!!20分~~

Consider a circle x^2 + y^2 + 2x + 6y + 5 = 0 and a point T(4 , 2) outside the circle.

(a) Find the centre C of the given circle.
x^2 + y^2 + 2x + 6y + 5 = 0
(x+1)^2+(y+3)^2=5
centre (-1,-3)

(b) Find the equation of the circle with the line segment TC as its diameter.
|TC|^2=(5^2+5^2)=50
Mid-point of TC=(3/2,-1/2)
So the equation of the circle is
(x-3/2)^2+(y+1/2)^2=50
x^2+y^2-3x+y+9/4+1/4=50
4x^2+4y^2-12x+4y-190=0
2x^2+2y^2-6x+2y-95=0

(c) If TP and TQ are tangents to the given circle and P and Q are the points of contact, find the equation of the line PQ.
Using formula
The equation of PQ is
4x + 2y + 2(4+x)/2 + 6(2+y)/2 + 5 = 0
4x + 2y + (4+x) + 3(2+y) + 5 = 0
5x+5y+15=0
x+y+3=0



2009-02-08 18:09:02 補充：
(B) 做錯
(x-3/2)^2+(y+1/2)^2=[√50/2]^2
(x-3/2)^2+(y+1/2)^2=50/4
(x-3/2)^2+(y+1/2)^2=25/2
x^2+y^2-3x+y+9/4+1/4=50/4
x^2+y^2-3x+y-10=0

2009-02-08 18:09:39 補充：
(C)不需用死力

2009-02-08 18:15:36 補充：
若果T(a,b)在圓外﹐而t相切圓於p,q﹐則pq方程為
(ax)+(by)+D(x+a)/2+E(y+b)/2+F= 0

2009-02-08 18:17:22 補充：
不過好難proof﹐你可以照下面那位
先求切線﹐再求切點﹐再求直線方程﹐不過出錯機會太大

http://pstock.idv.tw/files-pstock/9021003.pdf

General form of circle:
x^2 + y^2 + Dx + Ey + F = 0
or (x+D/2)^2 + (y+E/2)^2 = R^2
where centre of the circle is ( -D/2 , -E/2) and
radius, R = ( (-D/2)^2 + (-E/2)^2 ) ^(1/2)

Answers :
(a) The coordinates of center C are ( -1 , -3)

(b) Note that the mid-pt of TC is the centre of the circle,
Mid-pt is ( 3/2 , -1/2)
So, the equation of the circle with diameter TC
(x-3/2)^2 + (y+1/2)^2 = (4+1)^2 + (2+3)^2

(c)
以我理解 : P and Q are the points of contact mean that Tangets TP
and TQ to the circle at points P and Q on the circle respectively -->係

以下係我既方法(未必係快方法-_-)
To solve the problem ,
we need to find out the points P and Q .

General form of the tangent to the circle at point K (a,b) :
(ax)+(by)+D(x+a)/2+E(y+b)/2+F= 0

Let P and Q be (a,b) and (h,k) respectively ,
Note that tangent to the circle at K must be perpendicular to
the line KC which C is the centre of the circle
Consider P,
so, (2-b)/(4-a)+(b+3)/(a+1)= -1 --------(1)

Also, PC is radius of the circle
(b+3)^2+(a+1)^2 = 5 ----- (2)

From (1) and (2) , we have
a^2+b^2+2a+6b+5 = 0
a^2+b^2-3a+b-10 = 0
Solve the above equations,
we have a= 1 , b= - 4
a=- 2 , b= - 1
Note that there are two sets of solution because one set is for P
and the other is for Q.
Therefore,
slope of the PQ = (- 1 + 4) / (- 2 - 1)
Equation of PQ :
- 1 = (y + 4)/(x -1)
x+y+3 = 0