What are the radius and the center of the circle x^2-6x+y^2-2y=14 ?
Given the equation of the circle:
x^2 - 6x + y^2 - 2y = 14
We need to find the radius and the center of the circle.
Then, we need to rewrite into the standard form as follows:
(x-a)^2 + (y-b)^2 = r^2 such that (a,b) is the center and r is the radius.
To rewrite we need to complete the squares.
==> x^2 - 6x + 9 -9 + y^2 - 2y + 1 -1 = 14
==> (X-3)^2 + (y-1)^2 - 9 -1 = 14
==> (x-3)^2 + (y-1)^2 = 24
Then we conclude that:
The center is the point (3,1) and the radius is r=sqrt24= 2sqrt6
We have to pu the equation of the circle in this form:
(x − a)^2 + (y − b)^2 = r^2.
a and b are the coordinates of the center of circle and r is the radius.
Therefore, we'll create perfect squares in both x and y.
To complete the square in x, we'll add 9 both sides.
To complete the square in y, we'll add 1 both sides.(x^2 − 6x + 9) + (y^2 − 2y + 1) = 14 + 9 + 1 (x − 3)^2 + (y − 1)^2 = 24.
This circle has the radius r = 2sqrt6, and the coordinates of the center at (3, 1).