Find the equation of a circle passing through (3,7) and tangent to the lines x-3y+8=0 and y=3x.
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To find the equation of the circle passing through the point (3,7) and tangent to the lines x-3y+8=0 and y=3x, follow these steps:
Step 1: Find the center of the circle
Since the circle is tangent to the lines, the center of the circle must lie on the perpendicular bisector of the line segment joining the point (3,7) and the intersection point of the two lines. Let's call the intersection point (a,b).
To find the intersection point, we can solve the system of equations formed by the two given lines:
x - 3y + 8 = 0 ...(1)
y = 3x ...(2)
Substituting equation (2) into equation (1), we get:
x - 3(3x) + 8 = 0
x - 9x + 8 = 0
-8x + 8 = 0
-8x = -8
x = 1
Substituting x = 1 into equation (2), we get:
y = 3(1)
y = 3
So, the intersection point is (1, 3).
Now, let's find the midpoint of the line segment joining (3,7) and (1,3):
Midpoint x-coordinate = (3 + 1)/2 = 4/2 = 2
Midpoint y-coordinate = (7 + 3)/2 = 10/2 = 5
The center of the circle is (2, 5).
Step 2: Find the radius of the circle
The radius of the circle can be found by calculating the distance between the center (2, 5) and the given point (3, 7).
Using the distance formula:
Radius = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((3 - 2)^2 + (7 - 5)^2)
= sqrt(1^2 + 2^2)
= sqrt(5)
So, the radius of the circle is sqrt(5).
Step 3: Write the equation of the circle
The equation of a circle can be written in the standard form:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) represent the center coordinates and r represents the radius.
Plugging in the values we found earlier, the equation of the circle is:
(x - 2)^2 + (y - 5)^2 = (sqrt(5))^2
(x - 2)^2 + (y - 5)^2 = 5
Therefore, the equation of the circle passing through (3,7) and tangent to the lines x-3y+8=0 and y=3x is (x - 2)^2 + (y - 5)^2 = 5.