Find the center (h,k) and radius of the following circle.
x^2 + y^2 - 6x - 2y + 1 = 0
My answer is center= (-3,-1) and radius of 1. Am I correct?
x^2-6x +9 + y^2-2x+1=9 I added 9 to both sides.
(x-3)^2+(y-1)^2=3^2
I get center (3,1) and radius of 3
why 9?
To determine the center and radius of a circle, we need to rewrite the equation of the circle in the standard form:
(x - h)^2 + (y - k)^2 = r^2
So, let's complete the square to rewrite the given equation:
x^2 + y^2 - 6x - 2y + 1 = 0
First, rearrange the terms:
x^2 - 6x + y^2 - 2y = -1
Now, focus on the x terms. To complete the square, we need to take half of the coefficient of x (-6) and square it:
(-6/2)^2 = (-3)^2 = 9
Add this value to both sides of the equation:
x^2 - 6x + 9 + y^2 - 2y = -1 + 9
(x - 3)^2 + y^2 - 2y = 8
In the same way, complete the square for the y terms:
(-2/2)^2 = (-1)^2 = 1
Add this value to both sides of the equation:
(x - 3)^2 + (y - 1)^2 = 8 + 1
(x - 3)^2 + (y - 1)^2 = 9
Now we have the equation in standard form:
(x - h)^2 + (y - k)^2 = r^2
Comparing this to our equation, we can see that the center of the circle is (h, k) = (3, 1) and the radius squared is r^2 = 9.
Taking the square root of 9, we find the radius r = 3.
Therefore, the correct answer is the center (h, k) = (3, 1) and the radius r = 3.