2. Write the equation of the circle in standard form. Find the center, radius, intercepts, and graph the circle.
x2+y2+8x+2y+8=0
x^2 + y^2 + 8x + 2y + 8 = 0
x^2+8x+(8/2)^2+y^2+2y+(2/2)^2=-8+(8/2)^2
+(2/2)^2.
x^2+8x+16 + y^2+2y+1 = -8+16+1
Eq: (x+4)^2 + (y+1)^2 = 9
C(-4,-1).
r2 = 9
r = +-3
To write the equation of the circle in standard form, we need to complete the square in both x and y.
Step 1: Rearrange the terms in the equation to group the x-terms and y-terms together:
x^2 + 8x + y^2 + 2y + 8 = 0
Step 2: To complete the square for x, add half of the coefficient of x squared (which is 8/2 = 4) and square it. Add this value to both sides of the equation:
x^2 + 8x + 16 + y^2 + 2y + 8 = 16
(x + 4)^2 + y^2 + 2y + 8 = 16
Step 3: To complete the square for y, add half of the coefficient of y (which is 2/2 = 1) and square it. Add this value to both sides of the equation:
(x + 4)^2 + y^2 + 2y + 1 + 8 = 16 + 1
(x + 4)^2 + (y + 1)^2 + 9 = 17
Now the equation is in the standard form: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.
Comparing this to our equation, we can see that the center of the circle is (-4, -1) and the radius is the square root of 9, which is 3.
To find the intercepts, we set x = 0 and solve for y, and set y = 0 and solve for x:
Setting x = 0:
(0 + 4)^2 + (y + 1)^2 + 9 = 17
16 + (y + 1)^2 + 9 = 17
(y + 1)^2 = 17 - 16 - 9
(y + 1)^2 = -8
Since the square of a real number cannot be negative, there are no y-intercepts.
Setting y = 0:
(x + 4)^2 + (0 + 1)^2 + 9 = 17
(x + 4)^2 + 1 + 9 = 17
(x + 4)^2 = 17 - 1 - 9
(x + 4)^2 = 7
Taking the square root of both sides, we have:
x + 4 = ±√7
Solving for x, we get:
x = -4 ± √7
The x-intercepts are x = -4 + √7 and x = -4 - √7.
To graph the circle, plot the center (-4, -1), then use the radius of 3 to draw a circle around this point. The x-intercepts will be points on the circle that lie on the x-axis, and the y-intercepts will be points on the circle that lie on the y-axis.