Question

find the standard form of the equation for the circle with the following properties. center (1/6, - 1) and radius √10

Explanation:

Step1: Recall circle - equation formula

The standard - form 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 that the center is $(\frac{1}{6},-1)$ and the radius $r = \sqrt{10}$, so $h=\frac{1}{6}$, $k=-1$, and $r=\sqrt{10}$.

Step3: Substitute values into the formula

Substitute $h=\frac{1}{6}$, $k = - 1$, and $r=\sqrt{10}$ into the equation $(x - h)^2+(y - k)^2=r^2$. We get $(x-\frac{1}{6})^2+(y+1)^2 = (\sqrt{10})^2$.

Step4: Simplify the right - hand side

Since $(\sqrt{10})^2=10$, the standard form of the equation of the circle is $(x - \frac{1}{6})^2+(y + 1)^2=10$.

Answer:

$(x - \frac{1}{6})^2+(y + 1)^2=10$