How do you write the standard form of the equation of the circle whose diameter has endpoints of (-2, 4) and (4, 12)?

1 Answer

Answer:

#(x-1)^2+(y-8)^2=25#

Explanation:

The given data are the endpoints #E_1(x_1, y_1)=(-2, 4)# and #E_2(x_2, y_2)=(4, 12)# of the diameter #D# of the circle

Solve for the center #(h, k)#

#h=(x_1+x_2)/2=(-2+4)/2=1#
#k=(y_1+y_2)/2=(4+12)/2=8#

Center #(h, k)=(1, 8)#

Solve now for the radius #r#

#r=D/2=(sqrt((x_1-x_2)^2+(y_1-y_2)^2))/2#

#r=D/2=(sqrt((-2-4)^2+(4-12)^2))/2#

#r=D/2=sqrt(36+64)/2#

#r=D/2=sqrt(100)/2#

#r=D/2=10/2#

#r=5#

The standard form of the equation of the circle:

Center-Radius Form

#(x-h)^2+(y-k)^2=r^2#

#(x-1)^2+(y-8)^2=5^2#

#(x-1)^2+(y-8)^2=25#

God bless....I hope the explanation is useful.