A circle has the equation x squared plus y squared plus 8 x minus 8 y minus 49 equals 0
Graph the circle using the center (h,k) and radius r. Find the intercepts, if any, of the graph.
qaz
why all those cluttery words? This is math, not English.
x^2+y^2+8x-8y-49 = 0
x^2+8x + y^2-8y = 49
Now complete the squares:
x^2+8x+16 + y^2-8y+16 = 49+16+16
(x+4)^2 + (y-4)^2 = 81
I think you can prolly take it from there, eh?
To find and graph the circle using the given equation, we need to convert it into the standard form of a circle equation which is:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) represents the center of the circle and r represents the radius.
Given equation: x^2 + y^2 + 8x - 8y - 49 = 0
To convert this equation into the standard form, we need to complete the square for both the x and y terms.
Rearranging the equation, we get:
x^2 + 8x + y^2 - 8y = 49
Now, let's complete the square for the x terms. To do this, take half of the coefficient of x (which is 8), square it (which is 16), and add it to both sides of the equation. Similarly, complete the square for the y terms by taking half of the coefficient of y (which is -8), square it (which is 64), and add it to both sides of the equation.
x^2 + 8x + 16 + y^2 - 8y + 64 = 49 + 16 + 64
(x^2 + 8x + 16) + (y^2 - 8y + 64) = 129
Now, factor the x and y terms as perfect squares:
(x + 4)^2 + (y - 4)^2 = 129
Comparing this equation with the standard form equation, we can see that the center (h, k) is (-4, 4) and the radius (r) is the square root of 129.
So, the center of the circle is (-4, 4) and the radius is √129.
To graph the circle, plot the center point (-4, 4) on the coordinate plane. Then, draw a circle with a radius of √129 around this center point.
To find the intercepts of the graph, we need to find the coordinates where the circle intersects the x and y axes.
When a circle intersects the x-axis, the y-coordinate will be zero. So, we substitute y = 0 into the circle equation and solve for x:
(x + 4)^2 + (0 - 4)^2 = 129
(x + 4)^2 + 16 = 129
(x + 4)^2 = 113
x + 4 = ±√113
x = -4 ± √113
Therefore, the x-intercepts are (-4 + √113, 0) and (-4 - √113, 0).
Similarly, when a circle intersects the y-axis, the x-coordinate will be zero. So, we substitute x = 0 into the circle equation and solve for y:
(0 + 4)^2 + (y - 4)^2 = 129
16 + (y - 4)^2 = 129
(y - 4)^2 = 113
y - 4 = ±√113
y = 4 ± √113
Therefore, the y-intercepts are (0, 4 + √113) and (0, 4 - √113).
Now, you can plot the center, radius, and intercepts on the graph.