a. List twelve points, each with integer coordinates, that are 5 units from (-8, 1).

b. Find an equation of the circle containing these points.

Given that the coordinates are 5 units from (-8, 1).

We have to find the twelve points with integer coordinates.

$(-8,-6),(-8,-4),(-11,5),(-11,-3),(-12,4),(-12,-2),(-3,1),(-13,1),(-4,4),(-4,-2),(-5,5),(-5,-3)$The given point is (-8, 1),

We will add suitable points in order to get the points that are 5 units from (-8, 1).

First point:

$\begin{array}{l}{C}_1=\left(-8,1\right)+\left(0,5\right)\\ C_1=\left(-8,-6\right)\end{array}$

Second point:

$\begin{array}{l}{C}_2=\left(-8,1\right)+\left(0,-5\right)\\ C_2=\left(-8,-4\right)\end{array}$

Third point:

$\begin{array}{l}{C}_3=\left(-8,1\right)+\left(-3,4\right)\\ C_3=\left(-11,5\right)\end{array}$

Fourth point:

$\begin{array}{l}{C}_4=\left(-8,1\right)+\left(-3,-4\right)\\ C_4=\left(-11,-3\right)\end{array}$

Fifth point:

$\begin{array}{l}{C}_5=\left(-8,1\right)+\left(-4,3\right)\\ C_5=\left(-12,4\right)\end{array}$

Sixth point:

$\begin{array}{l}{C}_6=\left(-8,1\right)+\left(-4,-3\right)\\ C_6=\left(-12,-2\right)\end{array}$

Seventh point:

$\begin{array}{l}{C}_7=\left(-8,1\right)+\left(5,0\right)\\ C_7=\left(-3,1\right)\end{array}$

Eighth point:

$\begin{array}{l}{C}_8=\left(-8,1\right)+\left(-5,0\right)\\ C_8=\left(-13,1\right)\end{array}$

Nineth point:

$\begin{array}{l}{C}_9=\left(-8,1\right)+\left(4,3\right)\\ C_9=\left(-4,4\right)\end{array}$

Tenth point:

$\begin{array}{l}{C}_{10}=\left(-8,1\right)+\left(4,-3\right)\\ C_{10}=\left(-4,-2\right)\end{array}$

Eleventh point:

$\begin{array}{l}{C}_{11}=\left(-8,1\right)+\left(3,4\right)\\ C_{11}=\left(-5,5\right)\end{array}$

Twelfths point.

$\begin{array}{l}{C}_{12}=\left(-8,1\right)+\left(3,-4\right)\\ C_{12}=\left(-5,-3\right)\end{array}$

So, the twelve points from the point (-8, 1) are:

Given that the coordinates are 5 units from (-8, 1).

We have to write an equation of the circle containing these points.

Here, center = (h, k) = (-8, 1) and the radius = r = 5.

Plug the values in the standard form of a circle:

${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

$\begin{array}{l}{(x+8)}^2+{\left(y-1\right)}^2=5^2\\ {(x+8)}^2+{\left(y-1\right)}^2=25\end{array}$

So, the equation is${(x+8)}^2+{\left(y-1\right)}^2=25$