The equation of a circle, C, is x^2 + y^2 + 6x - 8y +9=0. Find the center of C and find the radius of C. Show that the C touches the x-axis. Find the equation of the circle which is a reflection of C in the y-axis
x^2+y^2+6x-8y+9 = 0
rearrange things a bit to get ready to complete the squares:
x^2+6x + y^2-8y = -9
x^2+6x+9 + y^2-8y+16 = -9+9+16
(x+3)^2 + (y-4)^2 = 16
so, the center is at (-3,4) with radius 4.
Note that the center is 4 units above the x-axis, so C touches the x-axis at (-3,0)
To reflect the circle in the y-axis, just flip the center to (+3,4). That makes the equation
(x-3)^2 + (y-4)^2 = 16
take a peek here:
http://www.wolframalpha.com/input/?i=plot+%28x%2B3%29^2+%2B+%28y-4%29^2+%3D+16%2C+%28x-3%29^2+%2B+%28y-4%29^2+%3D+16%2C+y%3D0%2C+x%3D0
To find the center of the circle, we need to complete the square for both x and y terms. The general form of a circle equation is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r is the radius.
Given equation: x^2 + y^2 + 6x - 8y + 9 = 0
Rearranging the terms, we have:
x^2 + 6x + y^2 - 8y = -9
To complete the square for x terms:
x^2 + 6x + 9 = -9 + 9
(x + 3)^2 = 0
To complete the square for y terms:
y^2 - 8y = 0
y^2 - 8y + 16 = 16
(y - 4)^2 = 16
Now we can rewrite the equation in the standard form of a circle:
(x + 3)^2 + (y - 4)^2 = 16
Comparing this with the general form, we have:
Center: (h, k) = (-3, 4)
Radius: r = √16 = 4
To show that the circle touches the x-axis, we can substitute y = 0 into the equation and see if it satisfies:
(x + 3)^2 + (0 - 4)^2 = 16
(x + 3)^2 + 16 = 16
(x + 3)^2 = 0
Since (x + 3)^2 = 0, it means that x + 3 = 0, so x = -3.
Thus, the point (-3, 0) lies on the circle, which demonstrates that the circle touches the x-axis.
To find the equation of the circle which is a reflection of C in the y-axis, we need to change the sign of the x-coordinate of the center.
Original center: (-3, 4)
Reflected center: (3, 4)
The equation of the reflected circle is:
(x - 3)^2 + (y - 4)^2 = 16
To find the center and radius of the circle equation, C: x^2 + y^2 + 6x - 8y + 9 = 0, we need to rewrite it in the standard form: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.
Step 1: Complete the square for x terms:
Start by rearranging the terms to group the x-terms together and the y-terms together:
(x^2 + 6x) + (y^2 - 8y) + 9 = 0
To complete the square for the x-terms, take half of the coefficient of x (which is 6), square it (which is 36), and add it to both sides of the equation:
(x^2 + 6x + 9) + (y^2 - 8y) = -9 + 9
(x + 3)^2 + (y^2 - 8y) = 0
Step 2: Complete the square for y terms:
Take half of the coefficient of y (which is -8), square it (which is 64), and add it to both sides of the equation:
(x + 3)^2 + (y^2 - 8y + 64) = 0 + 64
(x + 3)^2 + (y - 4)^2 = 64
Now the equation is in the standard form. Comparing it to the standard form, we can see that the center of the circle is (-3, 4) and the radius is the square root of 64, which is 8.
To show that the circle C touches the x-axis, we can check if the distance between the center of the circle (-3, 4) and the x-axis is equal to the radius (8). The distance between a point (x, y) and the x-axis is given by |y|.
In this case, the distance between (x, y) = (-3, 4) and the x-axis is |-4| = 4, which is equal to the radius (8). Therefore, the circle C touches the x-axis.
To find the equation of the circle that is a reflection of C in the y-axis, you simply change the sign of the x-coordinate of the center.
The center of C is (-3, 4), so the reflection of C in the y-axis has the center (3, 4). The radius remains the same (8).
Therefore, the equation of the reflected circle is (x - 3)^2 + (y - 4)^2 = 64.