If the endpoints of the diameter of a circle are (6, 3) and (2, 1), what is the standard form equation of the circle? a. (x + 4)2 + (y + 2)2 = 5 b. (x − 4)2 + (y − 2)2 = 5 c. (x + 4)2 + (y + 2)2 = 5 eliminate d. (x − 4)2 + (y − 2)2 = 5

(x - 4 )² + ( y - 2 )² = 5 → d

the equation of a circle in standard form is

(x - a )² + ( y - b)² = r²

where (a , b) are the coordinates of the centre and r is the radius

the centre is at the midpoint of the endpoints of the diameter and the radius is the distance from the centre to either of the 2 endpoints of the diameter.

to find the centre use the midpoint formula

(a , b) = [ (6 + 2) , (3 + 1 )] = (4 , 2)

to find r use the distance formula with points (4 , 2) and (6 ,3 )

r = √(6 - 4)² + ( 3 - 2)² = √( 4 + 1 ) = √5 ⇒ r² = 5

(x - 4)² + (y - 2)² = 5 is the equation

what are you doing do you need examples of variable questions?

step-by-step explanation:

5a-1=9               bring the one over the =

5a=10                     divide by your variable

a=2

brainliest? ?

here try this one

3a-24=0