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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

2 Answers

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(x-4)^2+(y-2)^2=5 =
User Schopenhauer
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(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) = [
(1)/(2) (6 + 2) ,
(1)/(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


User Kasperhj
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