a math ( circles)

Method A (d=r)
radius
=distance from centre to tangent
=|2+4-3|/√(1^2+1^2)
=3/√2
i.e. (x-2)^2+(y-4)^2=9/2 is the required equation

Method B (Δ=0)
Let the required equation be (x-2)^2+(y-4)^2=k
Put y=3-x
(x-2)^2+(x+1)^2=k
2x^2-2x+(5-k)=0
Δ=0
(-2)^2-4(2)(5-k)=0
4+8(k-5)=0
k-5=-1/2
k=9/2
i.e. (x-2)^2+(y-4)^2=9/2 is the required equation

Usually method A is quicker way,
but if sum of the roots is concerned, method B must be used.

2006-12-22 20:22:28 補充：
Point to note:Answer in general form or standard form are both accepted.i.e. You do not need to expand the expression (x-2)^2 (y-4)^2=9/2

x+y-3=0
slope=-1
so equation of line pass through centre is...
(y-4)/(x-2)=1 (as radius perpendicular to tangent)
y-4=x-2
x-y+2=0

then find the intersecting pt.
x+y-3=0 ---(1)
x-y+2=0 ---(2)
(1)+(2)
2x-1=0
x=1/2

(1)-(2)
2y-5=0
y=5/2
Ans: (1/2, 5/2)

radius=√[(2-1/2)^2+(4-5/2)^2]
=√[(3/2)^2+(3/2)^2]
=3/√2

equation of circle:
(x-2)^2+(y-4)^2=(3/√2)^2
x^2-2x+4+y^2-8y+16=9/2
x^2+y^2-2x-8y+31/2=0

2006-12-22 20:53:15 補充：
Sorry..typing mistake...equation of circle:(x-2)^2+(y-4)^2=(3/√2)^2x^2-4x+4+y^2-8y+16=9/2x^2+y^2-4x-8y+31/2=0