Find the equation for the circle with a diameter whose endpoints are (3,1) and (2,1)

hello :
an equation of the circle Center at the A(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : a = (3+2)/2 = 5/2   and  b = (1+1)/2 =1

let : B(3;1)    and ; C (2,1)

r = BC/2

BC =  √(3-2)²+(1-1)² )  =1

r = 1/2

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.