Part 2: Identify the center and radius of each. Then sketch the graph.

1. x^2 + y^2 - 4x + 6y + 9 = 0

2. x^2 + y^2 + 6x + 2y + 9 = 0

3. x^2 + y^2 - 6x + 2y + 1 = 0

4. x^2 + y^2 - 8x -2y + 8 = 0

See below for graphs

Step-by-step explanation:

I like to solve a set of problems like this once, then use a calculator or spreadsheet to crunch the numbers.

We can start with the general form of the equation for a circle:

  x^2 +y^2 +ax +by +c = 0

Completing the square gives ...

  (x^2 +ax +(a/2)^2) +(y^2 +by +(b/2)^2) = (a/2)^2 +(b/2)^2 -c

  (x +(a/2))^2 +(y +(b/2))^2 = (a/2)^2 +(b/2)^2 -c

Comparing this to the standard form formula ...

  (x -h)^2 + (y -k)^2 = r^2

we see the relations are ...

where (h, k) is the center and r is the radius.

Then your circle centers and radii are ...

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!