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

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

step-by-step explanation: