The graph of x2 −6x+9+y2 +2y+1 = 25 is a circle. Where is the center of the circle? What is the radius of the circle?
x^2−6x+9+y^2+2y+1 = 25
(x-3)^2 + (y+1)^2 = 5^2
Look familiar?
To find the center and radius of the circle represented by the equation x^2 - 6x + 9 + y^2 + 2y + 1 = 25, we can rewrite the equation in the standard form of a circle equation: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.
Let's begin by simplifying the given equation:
(x^2 - 6x + 9) + (y^2 + 2y + 1) = 25
(x^2 - 6x + y^2 + 2y) + (9 + 1 - 25) = 0
(x^2 - 6x + y^2 + 2y) - 15 = 0
Now, group the x and y terms together:
[(x^2 - 6x) + (y^2 + 2y)] - 15 = 0
(x^2 - 6x + 9) + (y^2 + 2y + 1) - 15 = 0
Notice that we can complete the squares for both the x-terms and the y-terms separately. To do this, we take half of the coefficient of the linear term, square it, and add it to both sides of the equation.
Let's complete the square for the x-terms:
(x^2 - 6x + 9) - 9 + (y^2 + 2y + 1) - 15 = -9 - 15
(x - 3)^2 + (y^2 + 2y + 1) - 9 - 15 = -24
(x - 3)^2 + (y^2 + 2y + 1) = -24 + 9 + 15
(x - 3)^2 + (y^2 + 2y + 1) = 0
Next, let's complete the square for the y-terms:
(x - 3)^2 + (y + 1)^2 = 0
Now we can see that the equation matches the standard form of a circle equation: (x - h)^2 + (y - k)^2 = r^2.
The center of the circle is represented by (h, k), therefore:
h = 3 (the x-coordinate of the center)
k = -1 (the y-coordinate of the center)
So, the center of the circle is (3, -1).
Since the equation is (x - 3)^2 + (y + 1)^2 = r^2, we can see that r^2 = 0. Taking the square root of both sides, we find that r = 0.
Therefore, the radius of the circle is 0.