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

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?

To find the equation of a circle passing through a specific point and tangent to two given lines, you can follow these steps:

Step 1: Find the center of the circle:
a) The center of the circle lies on the line that is the perpendicular bisector of the line segment joining the given point (3,7) and the center.
b) Find the equation of the line passing through the point (3,7) and perpendicular to the line x - 3y + 8 = 0.
c) Find the intersection point of the lines y = 3x and the line found in step 1b. That intersection point will be the center of the circle.

Step 2: Find the radius of the circle:
a) The distance between the center of the circle and the given point is equal to the radius of the circle. You can use the distance formula to find this distance.

Step 3: Write the equation of the circle:
a) Use the coordinates of the center obtained in step 1 and the radius obtained in step 2 to write the equation of the circle in the standard form: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center coordinates and r is the radius.

Let's now proceed with these steps to find the equation of the circle passing through (3,7) and tangent to the lines x-3y+8=0 and y=3x:

Step 1:
a) The line perpendicular to x - 3y + 8 = 0 passing through (3,7) has a slope equal to the negative reciprocal of the slope of the given line. The slope of the given line is 1/3, so the slope of the perpendicular line is -3.
b) Using the point-slope form of a line (y - y1) = m(x - x1), where (x1, y1) is the given point and m is the slope, we can write the equation of the perpendicular line as y - 7 = -3(x - 3).
Simplifying this equation, we get: y = -3x + 16.
c) Next, we need to find the intersection point of the perpendicular line y = -3x + 16 and the line y = 3x.
Solving these two equations simultaneously, we get:
-3x + 16 = 3x
6x = 16
x = 16/6 = 8/3
Substituting this value of x into either equation, we get y = 3 * (8/3) = 8.
Therefore, the center of the circle is (8/3, 8).

Step 2:
Using the distance formula, the distance between the center (8/3, 8) and the given point (3,7) is:
sqrt((8/3 - 3)^2 + (8 - 7)^2) = sqrt((8/3 - 9/3)^2 + 1^2) = sqrt((1/3)^2 + 1^2) = sqrt(1/9 + 1) = sqrt(10/9) = sqrt(10)/3.
Hence, the radius of the circle is sqrt(10)/3.

Step 3:
The equation of the circle is given by:
(x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center coordinates and r is the radius.
Substituting the values, we have:
(x - 8/3)^2 + (y - 8)^2 = (sqrt(10)/3)^2
Simplifying and expanding, the equation of the circle is:
(x - 8/3)^2 + (y - 8)^2 = 10/9