a circle has the equation
x^2+y^2+6x-6y-46=0
find the center and the radius of the euation. If there are any intercepts list them.
(x+3)^2 + (y-3)^2 = 64 = R^2
The origin is at x=-3 and y=3, and the radius is R=8
For the y intercept, set x=0 in the original equation and solve for y. For the x intercept, set y = 0 and solve for x.
To find the center and radius of a circle given its equation, we need to rewrite the equation in the standard form: (x - h)^2 + (y - k)^2 = r^2. In this form, the center of the circle is located at the point (h, k) and the radius is equal to r.
Let's start by completing the square for both the x and y terms in the equation:
x^2 + 6x + y^2 - 6y - 46 = 0
To complete the square for the x terms, we add (6/2)^2 = 9 to both sides:
x^2 + 6x + 9 + y^2 - 6y - 46 = 9
Simplifying, we get:
(x + 3)^2 + y^2 - 6y - 37 = 9
To complete the square for the y terms, we add (-6/2)^2 = 9 to both sides:
(x + 3)^2 + y^2 - 6y + 9 - 37 = 9 + 9
Simplifying further, we get:
(x + 3)^2 + (y - 3)^2 - 46 = 18
Rearranging the terms, we obtain:
(x + 3)^2 + (y - 3)^2 = 46 + 18
Now the equation is in the standard form: (x - h)^2 + (y - k)^2 = r^2. We can see that the center of the circle is at the point (-3, 3) and the radius is equal to the square root of 64, which is 8.
Therefore, the center of the circle is (-3, 3) and the radius is 8.
To find the intercepts, we set y = 0 and solve for x, and vice versa.
When y = 0, the equation becomes:
(x + 3)^2 + (0 - 3)^2 = 46 + 18
(x + 3)^2 + 9 = 64
(x + 3)^2 = 55
Taking the square root, we get:
x + 3 = ±√55
x = -3 ±√55
So the x-intercepts are approximately -3 + √55 and -3 - √55.
Similarly, when x = 0, the equation becomes:
(0 + 3)^2 + (y - 3)^2 = 46 + 18
9 + (y - 3)^2 = 64
(y - 3)^2 = 55
Taking the square root, we get:
y - 3 = ±√55
y = 3 ±√55
Hence, the y-intercepts are approximately 3 + √55 and 3 - √55.