Final answer:
The center of the circle represented by the equation x^(2)+y^(2)-6x-4y-12=0 is (3, 2) and the radius is 5.
Step-by-step explanation:
To determine the radius and center of the circle represented by the equation x^(2)+y^(2)-6x-4y-12=0, we need to rewrite the equation in the standard form for a circle, which is (x-h)^(2)+(y-k)^(2)=r^(2). In this form, the center of the circle is (h, k) and the radius is r.
First, we need to complete the square for both x and y terms. By adding and subtracting appropriate constants, we get (x^(2)-6x+9)+(y^(2)-4y+4)-9-4-12=0. Simplifying further, we have (x-3)^(2)+(y-2)^(2)-25=0.
Comparing this equation to the standard form, we can see that the center of the circle is (3, 2) and the radius is sqrt(25), which is equal to 5. Therefore, the center of the circle is (3, 2) and the radius is 5.