Dedric is trying to graph a circle on a computer software program that has a grid. The equation he is graphing is x2+y2−20x−14y+5=0 . Where would he place the center of the circle and what is its radius?

Answer 1

To determine the center and radius of the circle, we first need to rewrite the equation in standard form.

Completing the square for the x and y terms, we have:

(x^2 - 20x) + (y^2 - 14y) = -5

To complete the square for x, we take half of the coefficient of x, square it, and add it to both sides:

(x^2 - 20x + 100) + (y^2 - 14y) = -5 + 100
(x - 10)^2 + (y^2 - 14y) = 95

Now we complete the square for y:

(x - 10)^2 + (y^2 - 14y + 49) = 95 + 49
(x - 10)^2 + (y - 7)^2 = 144

Comparing this to the standard form of a circle equation, (x - h)^2 + (y - k)^2 = r^2, we can see that the center of the circle is at (h, k) = (10, 7) and the radius r = √144 = 12.

Therefore, Dedric would place the center of the circle at (10, 7) on the grid software program and the radius would be 12 units.