Posted by xavier on Friday, January 25, 2013 at 9:36am.
let the centre be C(a,b)
then the distance from C to x-3y+8 = 0 is
|a - 3b+8|/√10
then the distance from C to 3x-y=0 is
|3a - b|/√10
so 3a-b = a-3b + 8 or 3a-b = -a + 3b - 8
2a + 2b = 8 or 4a - 4b = -8
a + b = 4 or or a - b = -2
let's use a-b=-2 or b = a+2 , more reasonable from my sketch
also radius = √(a-3)^2 + (b-7)^2
so
(3a-b)/√10 = √(a-3)^2 + (b-7)^2
square both sides:
(3a - b)^2 /10 = (a-3)^2 + (b-7)^2
but b = a+2
((3a - a-2)^2) /10 = (a-3)^2 + (a-5)^2
4a^2 - 8a + 4 = 10(a^2 - 6a + 9 + a^2 - 10a + 25)
4a^2 - 8a + 4 = 20a^2 - 160a + 340
16a^2 - 152a + 336 = 0
2a^2 - 19a + 42 = 0
(a-6)(2a-7) = 0
a = 6 or a = 7/2
b = 8 or b = 11/2
so we have to possible centres (6,8) and (3.5 , 5.5)
each one must be below y = 3x and above x-3y+8=0
or y < 3x and y > (x+8)/3
(a quick arithmetic check shows both are possible)
I will use the (6,8)
so a possible equation is
(x-6)^2 + (y-8)^2 = r^2
but (3,7) lies on it, so
(-3)^2 + (-1)^2 = r^2 = 10
(x-6)^2 + (y-8)^2 = 10 is such a circle.
thank you very much!
the two lines intersect at (1,3)
the two lines have slope 1/3 and 3.
So, the center of the circle lies on the line y=x+2. (why?)
the distance from (h,k) to ax+by+c=0 is
|ah+bk+c|/√(a^2+b^2), so if our circle has center (h,k), its radius is
|h-3k+8|/√10 or |3h-k|/√10
Since k=h+2,
|h-3h-6+8| = |-2h+2|
|3h-(h+2)| = |2h-2|
so the distance to the two lines is the same. (whew)
So, our circle is
(x-h)^2 + (y-(h+2))^2 = (2h-2)^2/10
Now, we know that (3,7) is on the circle, so
(3-h)^2 + (7-h-2)^2 = 4(h-2)^2/10
5(3-h)^2 + 5(5-h)^2 = 2(h-2)^2
45-30h+5h^2 + 125-50h+5h^2 = 2h^2-8h+8
8h^2 - 72h + 162 = 0
4h^2 - 36h + 81 = 0
(2h-9)^2 = 0
h = 9/2
So, the circle is
(x-9/2)^2 + (y-(9/2+2))^2 = (18/2-2)^2/10
(x-9/2)^2 + (y-13/2)^2 = 49/10
Reiny and I took basically the same approach, but we arrived at different circles.
Both are correct, but the point (3,7) lies on opposite sides of the radius to the tangent.
Cool!
Are you all using directed distance?