The graph of x^2 + y^2 = 6x + 81 is the graph of a(n) _______.
Circle?
correct
Thank you
Yes, the graph of the equation x^2 + y^2 = 6x + 81 represents a circle.
To determine whether an equation represents a circle, we need to examine its form. A general form of a circle equation is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle, and r represents the radius.
In the given equation, x^2 + y^2 = 6x + 81, we can rearrange it to resemble the general form of a circle equation. By subtracting 6x and 81 from both sides, we get:
x^2 - 6x + y^2 = 81
Now, to complete the square, we need to add and subtract (-6/2)^2 = 9 to the left side of the equation:
x^2 - 6x + 9 + y^2 = 81 + 9
This can be further simplified as:
(x - 3)^2 + y^2 = 90
Comparing this equation with the general form of a circle equation, we can identify that the center of the circle is (3, 0) and the radius is √90 or approximately 9.49.
Therefore, the graph of x^2 + y^2 = 6x + 81 is the graph of a circle centered at (3, 0) with a radius of approximately 9.49 units.