Final Answer:
The center of the circle is (3, -8) and the radius is 7.
Step-by-step explanation:
To find the center and radius of a circle given its equation in standard form, we follow these steps:
1. Isolate the term containing x and y variables on one side of the equation.
2. Set this expression equal to r^2, where r is the radius.
3. Solve for x and y in terms of r.
4. The center is located at (x, y).
Let's apply these steps to our given equation:
(x-3)^2 + (y+8)^2 = 39
1. Isolate the term containing x and y variables:
(x-3)^2 + (y+8)^2 = 39
(x-3)^2 = 39 - (y+8)^2
(x-3)^2 = 13^2 - 2(13)(y+8) + (y+8)^2
(x-3)^2 = 169 - 276y - 504y^2 + 576y^3
2. Set this expression equal to r^2:
(x-3)^2 = r^2
(x-3) = ±r (taking square root on both sides)
x = 3 ± r (solving for x)
3. Solve for y in terms of x and r:
(y+8) = ±r (taking square root on both sides)
y = -8 ± r (solving for y)
4. The center is located at (x, y):
The center is located at (3 ± r, -8 ± r). However, since we want a unique center, we'll choose the one with a positive value for r. Therefore, the center is located at (3 + r, -8 - r), which simplifies to (3, -8). This is our final answer.