Question
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?
Answers
Answered by
GPT 3.5
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.
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.