The equation of a circle is x^2 + 6x + y^2 - 2y = 15. What are the center and radius of the circle? Please show work.
Complete the squares for x and y to put into standard (x - h)^2 + (y - k)^2 = r^2
for center (h , k) and rdius r
x^2 + 6x + y^2 - 2y = 15
x^2 + 6x + 9 + y^2 - 2y + 1 = 15 + 9 + 1 = 25
(x + 3)^2 + (y - 1)^2 = 25
comparing to the standard form:
center (h , k) is (-3 , 1)
r^2 = 25 > radius r = 5
Center (-3,1)
xc= -3 yc=1 re5
To find the center and radius of a circle, we need to rewrite the equation of the circle into the standard form: (x - h)^2 + (y - k)^2 = r^2.
Let's complete the square for both the x and y terms in the given equation.
1. Start with the given equation: x^2 + 6x + y^2 - 2y = 15.
2. Group the x terms and complete the square:
(x^2 + 6x) + (y^2 - 2y) = 15.
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 + 36) + (y^2 - 2y) = 15 + 36.
(x + 3)^2 + (y^2 - 2y) = 51.
3. Group the y terms and complete the square:
(x + 3)^2 + (y^2 - 2y) = 51.
To complete the square for the y terms, take half of the coefficient of y (which is -2), square it (which is 4), and add it to both sides of the equation:
(x + 3)^2 + (y^2 - 2y + 4) = 51 + 4.
(x + 3)^2 + (y - 1)^2 = 55.
Now we have the equation in the standard form, where (h, k) is the center of the circle and r is the radius.
Comparing it with the standard form: (x - h)^2 + (y - k)^2 = r^2, we can identify that the center of the circle is (-3, 1) and the radius is the square root of 55.
So, the center of the circle is (-3, 1) and the radius is √55.