Determine whether the given equation is a circle, a point, or a contradiction (no real graph).
x^2 + y^2 - 2 x - 16 y + 52 = 0 .
The equation is a
. (Enter "circle", "point", or "contradiction" in the answer space without quotes.)
To determine the type of graph represented by the given equation, we need to analyze its form.
The equation is in the form:
x^2 + y^2 - 2x - 16y + 52 = 0
This equation represents a standard form of a circle. In general, a circle equation can be written in the form:
(x - h)^2 + (y - k)^2 = r^2
Comparing this general form to the given equation, we can see that the given equation can be rearranged to match this form:
(x^2 - 2x) + (y^2 - 16y) = -52
(x^2 - 2x + 1) + (y^2 - 16y + 64) = -52 + 1 + 64
(x - 1)^2 + (y - 8)^2 = 13
The equation is now in the form (x - h)^2 + (y - k)^2 = r^2, where the center of the circle is at the point (h, k) = (1, 8) and the radius of the circle is sqrt(r^2) = sqrt(13).
Therefore, the given equation represents a circle.