What is the center of the circle with the equation x2 + y2 + 6x - 8y - 11 = 0?Question 10Select one:a.(3, 4)b.(-3, 4)c.(4, -3)d.(6, 0)

The equation of a circle is given by (x-h)² + (y-k)² = r² where (h,k) is the center of the circle and r is the radius.

The given equation is x² + y² + 6x - 8y - 11 = 0.

To find the center of the circle, we need to complete the square for the x and y terms.

Rearrange the equation to group the x terms and y terms together:

(x² + 6x) + (y² - 8y) = 11

To complete the square, we take half of the coefficient of x, square it and add it to both sides. Half of 6 is 3, and 3 squared is 9.

Similarly, we take half of the coefficient of y, square it and add it to both sides. Half of -8 is -4, and -4 squared is 16.

So, we add 9 and 16 to both sides of the equation:

(x² + 6x + 9) + (y² - 8y + 16) = 11 + 9 + 16

This simplifies to:

(x + 3)² + (y - 4)² = 36

So, the center of the circle is at (-3, 4), which corresponds to option b.(-3, 4).