Question

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

Answer 1

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`

Answer 2

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`