Question

write the standard form of the equation of the circle with its center at (-3,0), and a radius of 6. what is the equation of the circle in standard form? (simplify your answer.)

Explanation:

Step1: Recall circle - standard - form formula

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

Step2: Identify the values of $h$, $k$, and $r$

Given that the center is $(-3,0)$, so $h=-3$, $k = 0$, and the radius $r = 6$.

Step3: Substitute the values into the formula

Substitute $h=-3$, $k = 0$, and $r = 6$ into $(x - h)^2+(y - k)^2=r^2$. We get $(x-(-3))^2+(y - 0)^2=6^2$.

Step4: Simplify the equation

$(x + 3)^2+y^2=36$.

Answer:

$(x + 3)^2+y^2=36$