what is the radius of the circle described by the equation $(x - 6)^2+(y - 4)^2-64 = 0$?\na 64\nb 16\nc…

Explanation:

Step1: Rewrite the equation in standard form

The standard form of a circle equation is $(x - a)^2+(y - b)^2=r^2$. Given $(x - 6)^2+(y - 4)^2-64 = 0$, we can rewrite it as $(x - 6)^2+(y - 4)^2=64$.

Step2: Identify the radius

Comparing $(x - 6)^2+(y - 4)^2=64$ with $(x - a)^2+(y - b)^2=r^2$, we have $r^2 = 64$. Taking the square - root of both sides, since $r>0$, we get $r = 8$.

Answer:

D. 8