Amaths - Circle

Let the centre of the circle be (h, k). Then it should be equidistant from both of the given lines, i.e.
|(3h - k + 3)/√(32 + 12)| = |(h - 3k - 7)/√(12 + 32)|
|3h - k + 3| = |h - 3k - 7|
3h - k + 3 = h - 3k - 7 or 3h - k + 3 = -h + 3k + 7
2h = -2k - 10 or 4h = 4k + 4
h = - k - 5 or k + 1
Therefore the radius of the circle can be expressed as:
|3h - k + 3|/√10 = √[(h - 2)2 + (k + 1)2]
(3h - k + 3)2/10 = (h - 2)2 + (k + 1)2
For h = k + 1:
(2h + 4)2/10 = (h - 2)2 + h2
4h2 + 16h + 16 = 10(2h2 - 4h + 4)
h2 + 4h + 4 = 5(h2 - 2h + 2)
h2 + 4h + 4 = 5h2 - 10h + 10
4h2 - 14h + 6 = 0
2h2 - 7h + 3 = 0
(2h - 1)(h - 3) = 0
h = 1/2 or 3
k = -1/2 or 2
So the possible centres are (1/2, -1/2) and (3, 2) and respectively, their radii are (√10)/2 and √10
Hence the possible equations are:
(x - 1/2)2 + (y + 1/2)2 = 10/4
x2 + y2 - x - y + 1/2 = 10/4
4x2 + 4y2 - 4x - 4y - 8 = 0
OR
(x - 3)2 + (y - 2)2 = 10
x2 + y2 - 6x - 4y + 13 = 10
x2 + y2 - 6x - 4y + 3 = 0

2008-07-06 19:44:45 補充：
For h = - k - 5, the discriminant of the quadratic equation formed is negative, i.e. no real solutions.

第二位回答者也答得很詳盡,給我另一條思路,很有新意.謝謝你!

Equation of the angle bisector of the 2 lines are:
3x-y+3/[sqrt(3^2 + (-1)^2) = +/- {x-3y-7]/[sqrt(1^2 + (-3)^2]
3x - y + 3 = +/- (x-3y -7).That is:
3x - y + 3 = x -3y -7, 2x + 2y +10 = 0, x + y +5 = 0 or
3x - y + 3 = -x +3y + 7 , 4x -4y -4 = 0 , x - y -1 = 0.
By sketching the 2 given lines and point (2,-1), the angle bisector with positive slope should be the one to take in order for the circles to exist. That is x-y-1 = 0.
Since the circle touches both lines, by property of tangent, centre of circle must be on this angle bisector. Let h be the x-coordinate of the centre, the y-coordinate will be h-1. That is the centre is [h, (h-1)].
Distance from this point to one of the lines = distance from this point to (2,-1) = radius of circle. That is,
[3h - (h-1) +3]/{sqrt[3^2 +(-1)^2]} = sqrt[(h-2)^2 + (h-1+1)^2]
(2h + 4)^2 = 10(h^2 + 4 - 4h + h^2)
4(h+2)^2 = 10(2h^2 -4h +4)
(h+2)^2 = 5(h^2 - 2h + 2)
h^2 + 4h + 4 = 5h^2 -10h + 10
4h^2 - 14h +6 = 0
2h^2 -7h + 3 = 0
(2h -1)(h-3) = 0
that is h = 1/2 or 3. So centre of circle are (1/2, -1/2) and (3, 2).
Radius = sqrt(9/4 + 1/4) =1/2 sqrt10 and sqrt(1 + 9) = sqrt10.
So equations of circle are:
(x-1/2)^2 + (y+1/2)^2 = 10/4. And
(x-3)^2 + (y-2)^2 = 10. Or
x^2 + y^2 -x +y -2 = 0 and
x^2 +y^2 -6x -4y +3 = 0.

唔明你問乜,要搵兩點?