Circle

Let the point(s) of contact be (-1+4t,2-3t). (parametric eq. of 3x+4y=5)
The circle can be expressed as
(x+1- 4t)^2+(y-2+3t)^2+k(3x+4y-5)=0
Note: generate by the "circle" (x+1- 4t)^2+(y-2+3t)^2=0 and 3x+4y=5

Passes through (1, 3) => (2- 4t)^2+(1+3t)^2+10k=0
Passes through (2, 1) => (3- 4t)^2+(3t-1)^2+ 5k=0
=> (2- 4t)^2+(1+3t)^2= 2[(3- 4t)^2+(3t-1)^2]
25t^2-50t+15=0 , t= 1+/- √(2/5)
=> the tangent points are (-1+4t, 2-3t) =
( 3+4√(2/5), -1-3√(2/5) ) and ( 3- 4√(2/5), -1+3√(2/5) )

2009-08-23 19:22:38 補充：
method2: line AB(2x+y=5) meets 3x+4y=5 at P(3,-1)
PA*PB=10= PQ^2 =>contact points Q = (3, -1) +/- √10 *(4, -3)/5

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Let the point of contact be P(h,k).
The line joining points A(1,3) and B(2,1) meets the tangent at Q.
3x + 4y = 5 . . . . . . (1)
Equation of AB is :
( y – 3 ) / ( x – 1 ) = ( 3 – 1 ) / ( 1 – 2 )
y – 3 = –2x + 2
y = 5 – 2x . . . . . (2)

2009-08-23 18:43:45 補充：
Substitute (2) into (1) : 3x + 20 – 8x = 5
5x = 15
x = 3
y = –1
The line joining points A(1,3) and B(2,1) meets the tangent at Q(3, –1).

2009-08-23 18:44:13 補充：
QA =√[ ( 3 – 1 )^2 + ( –1 – 3 )^2 ] =√( 4 + 16 ) =√20 = 2√5
QB =√[ ( 3 – 2 )^2 + ( –1 – 1 )^2 ] =√( 1 + 4 ) =√5
QP =√[ ( h – 3 )^2 + ( k + 1 )^2 ] =√( h^2 + k^2 – 6h + 2k + 10 )
△APQ and △PBQ are similar.
QP / QA = QB / QP
QP^2 = QA X QB

2009-08-23 18:44:37 補充：
h^2 + k^2 – 6h + 2k + 10 = 2√5 X√5
h^2 + k^2 – 6h + 2k = 0 . . . . . (3)
P(h,k) lies on (1).
3h + 4k = 5
k = ( 5 – 3h ) / 4 . . . . . . . . (4)
You can obtain the point of contact by solving (3) and (4).

2009-08-23 19:09:52 補充：
Substitute (4) into (3) :
h^2 + [ ( 5 – 3h ) / 4 ]^2 – 6h + 2 ( 5 – 3h ) / 4 = 0
5h^2 – 30h + 13 = 0
h = 3 +/- 4√10 / 5
h = 0.470 or 5.530
k = –1 -/+ 3√10 / 5
k = 0.897 or -2.897
The point of contact is (0.470,0.897) or (5.530,-2.897).