How would you determine the equation of the circle which passes through the points D(-5,-5), E(-5,15), F(15,15)?

2 Answers
Jan 10, 2016

Answer:

Substitute each point to the equation of the circle, develop 3 equations, and substract the ones that have at least 1 coordinate common (#x# or #y#).

Answer is:

#(x-5)^2+(y-5)^2=200#

Explanation:

The equation of the circle:

#(x-α)^2+(y-β)^2=ρ^2#

Where #α# #β# are the coordinates of the center of the circle.

Substitute for each given point:

Point D

#(-5-α)^2+(-5-β)^2=ρ^2#

#(-(5+α))^2+(-(5+β))^2=ρ^2#

#(5+α)^2+(5+β)^2=ρ^2#

#5^2+2*5α+α^2+5^2+2*5β+β^2=ρ^2#

#α^2+β^2+10α+10β+50=ρ^2# (Equation 1)

Point E

#(-5-α)^2+(15-β)^2=ρ^2#

#(5+α)^2+(15-β)^2=ρ^2#

#5^2+2*5α+α^2+15^2-2*15β+β^2=ρ^2#

#α^2+β^2+10α-30β+250=ρ^2# (Equation 2)

Point F

#(15-α)^2+(15-β)^2=ρ^2#

#15^2-2*15α+α^2+15^2-2*15β+β^2=ρ^2#

#α^2+β^2-30α-30β+450=ρ^2# (Equation 3)

Substract equations #(1)-(2)#

#α^2+β^2+10α+10β+50=ρ^2#
#α^2+β^2+10α-30β+250=ρ^2#

#40β-200=0#

#β=200/40#

#β=5#

Substract equations #(2)-(3)#

#α^2+β^2+10α-30β+250=ρ^2#
#α^2+β^2-30α-30β+450=ρ^2#

#40α-200=0#

#α=200/40#

#α=5#

Now that #α# and #β# are known, substitute them in any of the points (we will use point #D(-5,-5)#):

#(x-α)^2+(y-β)^2=ρ^2#

#(-5-5)^2+(-5-5)^2=ρ^2#

#(-10)^2+(-10)^2=ρ^2#

#2(-10)^2=ρ^2#

#ρ^2=200#

So the equation of the circle becomes:

#α=5#
#β=5#
#ρ^2=200#

#(x-α)^2+(y-β)^2=ρ^2#

#(x-5)^2+(y-5)^2=200#

Jan 10, 2016

Answer:

The circle's equation is #(x-5)^2+(y-5)^2=200#

Explanation:

First we need to find the equation of two lines, each one perpendicular to the segments formed by a pair of the given points and passing through the midpoint of this pair of points.
Since points D and E (#x_D=x_E=-5#) are in a line parallel to the axis-Y (#x=0#) and points E and F (#y_E=y_F=15#) are in a line parallel to the axis-X (#y=0#) it's convenient to choose these pairs of points.

Equation of Line DE, where #x_D=x_E=-5#
#x=-5#
Equation of line 1 perpendicular to DE and passing through midpoint #M_(DE)#
#M_(DE) ((x_D+x_E)/2,(y_D+y_E)/2)# => #M_DE(-5, 5)#
line 1#-> y=5#

Equation of Line EF, where #y_E=y_F=15#
#y=15#
Equation of line 2 perpendicular to EF and passing through midpoint #M_(EF)#
#M_(EF) ((x_E+x_F)/2,(y_E+y_F)/2)# => #M_EF(5,15)#
line 2#->x=5#

Combining equations of lines 1 and 2 (#y=5# and #x=5#) we find the circle's center, point C
#C(5,5)#

The distance between point C to any of the given points is equal to the circle's radius
#R=d_(CD)=sqrt((-5-5)^2+(-5-5)^2)=sqrt(100+100)=sqrt(200)#

In the formula of the equation of the circle:
#(x-x_C)^2+(y-y_C)^2=R^2#
#(x-5)^2+(y-5)^2=200#