Find the equation of the circle that passes through (2,2) and tangent to the line x=1 and x=6.
the two lines are 5 apart, so the radius is 5/2, with the center at (7/2,k) so
(x - 7/2)^ + (y-k)^2 = (5/2)^2
The point (2,2) is on the line, so
(2 - 7/2)^2 + (2-k)^2 = (5/2)^2
k = 0 or 4
(x - 7/2)^2 + (y-4)^2 = (5/2)^2
(x - 7/2)^2 + y^2 = (5/2)^2
So, the question is misleading. There are two circles that fit the conditions:
http://www.wolframalpha.com/input/?i=plot+%28x+-+7%2F2%29^2+%2B+%28y-4%29^2+%3D+%285%2F2%29^2%2C+%28x+-+7%2F2%29^2+%2B+y^2+%3D+%285%2F2%29^2+and+x%3D1+and+x%3D6
Why did the circle go to the therapist?
Because it felt like it was being tangentially touchy with the lines!
But don't worry, let's solve this problem and get the equation of the circle.
First, we know that the circle passes through (2,2) and is tangent to the vertical lines x=1 and x=6.
The equation of a circle can generally be written as (x - a)^2 + (y - b)^2 = r^2, where (a,b) is the center of the circle and r is the radius.
Since our circle is tangent to the vertical lines x=1 and x=6, its center must lie on the line halfway between them, which is x=3.5.
So, the x-coordinate of the center is 3.5.
Since the circle passes through (2,2), we can substitute these values into the equation and solve for the y-coordinate of the center:
(2 - 3.5)^2 + (2 - b)^2 = r^2
(1.5)^2 + (2 - b)^2 = r^2
2.25 + (2 - b)^2 = r^2
Now, we just need to find the radius (r). Since the circle is tangent to the lines x=1 and x=6, the radius will be the distance between the center of the circle and either of these lines. In this case, it will be the distance from x=3.5 to x=1 (or x=6), which is 2.5 units.
So, the radius (r) is 2.5 units.
Substituting this value into the equation, we have:
2.25 + (2 - b)^2 = (2.5)^2
2.25 + (2 - b)^2 = 6.25
(2 - b)^2 = 6.25 - 2.25
(2 - b)^2 = 4
Taking the square root of both sides, we get:
2 - b = 2 or 2 - b = -2
If 2 - b = 2, then b = 0
If 2 - b = -2, then b = 4
Therefore, the center of the circle can be (3.5, 0) or (3.5, 4).
Now we can write the equation of the circle:
For the center (3.5, 0):
(x - 3.5)^2 + (y - 0)^2 = (2.5)^2
(x - 3.5)^2 + y^2 = 6.25
For the center (3.5, 4):
(x - 3.5)^2 + (y - 4)^2 = (2.5)^2
(x - 3.5)^2 + (y - 4)^2 = 6.25
And there you have it! The equation of the circle that passes through (2,2) and is tangent to the lines x=1 and x=6.
To find the equation of the circle, we need to find the center and the radius.
Step 1: Find the center of the circle.
Since the circle passes through the point (2,2) and is tangent to the line x=1 and x=6, the center of the circle must lie on the line x=3 (midpoint between 1 and 6). Therefore, the x-coordinate of the center is 3.
Step 2: Find the y-coordinate of the center.
Since the circle is tangent to the line x=1 and x=6, the radius is the perpendicular distance from the center to either of these lines. Since the circle passes through the point (2,2), the perpendicular distance is the difference between the y-coordinate of the center and 2. This means the y-coordinate of the center is 2.
Step 3: Find the radius of the circle.
The radius is the distance from the center to any point on the circle. Since the circle passes through (2,2), we can use the distance formula to find the radius:
r = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((3 - 2)^2 + (2 - 2)^2)
= sqrt(1 + 0)
= sqrt(1)
= 1
Step 4: Write the equation of the circle.
Using the center (3,2) and the radius of 1, the equation of the circle is:
(x - 3)^2 + (y - 2)^2 = 1
Therefore, the equation of the circle is (x - 3)^2 + (y - 2)^2 = 1.
To find the equation of the circle that passes through the point (2, 2) and is tangent to the lines x = 1 and x = 6, we can follow these steps:
Step 1: Find the center of the circle
Since the circle is tangent to the lines x = 1 and x = 6, the center of the circle lies on the line connecting the midpoints of the two lines.
The midpoint between x = 1 and x = 6 is found by taking the average of their x-coordinates:
Midpoint_x = (1 + 6) / 2 = 7 / 2 = 3.5
Since the circle is tangent to the line x = 1, the x-coordinate of its center is 1 unit away from x = 1. Therefore, the x-coordinate of the center is: Center_x = 1 + 1 = 2
So, the center of the circle is (2, Center_y), where Center_y is the y-coordinate of the center.
Step 2: Find the radius of the circle
The radius of the circle is the distance between the center and the point on the circle (2, 2).
The distance formula between two points (x1, y1) and (x2, y2) is given by:
r = √[(x2 - x1)^2 + (y2 - y1)^2]
In this case, (x1, y1) = (2, Center_y) and (x2, y2) = (2, 2).
So we have: r = √[(2 - 2)^2 + (2 - Center_y)^2] = √[(2 - Center_y)^2]
Step 3: Substitute the center and radius into the equation of a circle
The general equation of a circle with center (h, k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2
Substituting the center (2, Center_y) and radius √[(2 - Center_y)^2], we get:
(x - 2)^2 + (y - Center_y)^2 = (2 - Center_y)^2
Therefore, the equation of the circle is (x - 2)^2 + (y - Center_y)^2 = (2 - Center_y)^2.
Remember to substitute the value of Center_y you obtain from the midpoint calculation in the previous step to get the final equation.