Question

Find the equation of a circle passing through (3,7) and tangent to the lines x-3y+8=0 and y=3x.

Please help me

Answer 1

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.

Answer 2

thank you very much!

Answer 3

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

Answer 4

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!

Answer 5

Are you all using directed distance?