Question

write the standard form of the equation of the circle with the given center and radius. center (3, - 1), r = \sqrt{3}
the equation of the circle in standard form is (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 values of $h$, $k$, and $r$

Given center $(3,-1)$ and $r = \sqrt{3}$, so $h = 3$, $k=-1$, and $r=\sqrt{3}$.

Step3: Substitute values into the formula

Substitute $h = 3$, $k=-1$, and $r=\sqrt{3}$ into $(x - h)^2+(y - k)^2=r^2$. We get $(x - 3)^2+(y+1)^2=(\sqrt{3})^2$.

Step4: Simplify the right - hand side

Since $(\sqrt{3})^2 = 3$, the equation of the circle in standard form is $(x - 3)^2+(y + 1)^2=3$.

Answer:

$(x - 3)^2+(y + 1)^2=3$