Answer:
A: (-6,-5)
B: r = 3
Explanation:
To find the center and radius of the circle given by the equation x^2 + y^2 + 12x + 10y + 52 = 0, we first need to rewrite the equation in standard form, which is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is its radius.
Part A:
To rewrite the equation in standard form, we need to complete the square for both x and y. We can do this by adding and subtracting the square of half the coefficient of x and y respectively, as follows:
x^2 + 12x + y^2 + 10y + 52 = 0
(x^2 + 12x + 36) + (y^2 + 10y + 25) = -52 + 36 + 25
(x + 6)^2 + (y + 5)^2 = 9
Now we can see that the equation is in standard form, with center (-6, -5) and radius 3. Therefore, the center point of the circle is (-6, -5).
Part B:
The radius of the circle is given by the square root of the right-hand side of the standard form equation, which is 3. Therefore, the radius of the circle is 3.