# What are the radius and the center of the circle x^2-6x+y^2-2y=14 ?

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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).**