Question

determine the equation of a circle with radius 6 and its center at the origin.
a $\frac{x^{2}}{12}+\frac{y^{2}}{6}=1$
b $x^{2}+y^{2}=36$
c $x^{2}+y^{2}=12$
d $x^{2}+y^{2}=6$

Explanation:

Step1: Recall circle - equation formula

The standard equation of a circle with center $(h,k)$ and radius $r$ is $(x - h)^{2}+(y - k)^{2}=r^{2}$.

Step2: Identify center and radius values

The center is at the origin $(0,0)$, so $h = 0$ and $k = 0$, and $r=6$.

Step3: Substitute values into formula

Substituting $h = 0$, $k = 0$, and $r = 6$ into $(x - h)^{2}+(y - k)^{2}=r^{2}$, we get $(x-0)^{2}+(y - 0)^{2}=6^{2}$, which simplifies to $x^{2}+y^{2}=36$.

Answer:

B. $x^{2}+y^{2}=36$