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

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

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