Question

what is the center of a circle represented by the equation $(x-5)^2+(y+6)^2=4^2$?$(-6,5)$$(-5,6)$$(5,-6)$$(6,-5)$

Explanation:

Step1: Recall circle standard form

The standard equation of a circle is $(x-h)^2+(y-k)^2=r^2$, where $(h,k)$ is the center.

Step2: Match given equation to standard form

Given equation: $(x-5)^2+(y+6)^2=4^2$, rewrite $y+6$ as $y-(-6)$. So $h=5$, $k=-6$.

Answer:

(5, -6)