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

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 $y$).

${\left(x - 5\right)}^{2} + {\left(y - 5\right)}^{2} = 200$

#### 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 $\left(1\right) - \left(2\right)$

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 $\left(2\right) - \left(3\right)$

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 $D \left(- 5 , - 5\right)$):

(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

${\left(x - 5\right)}^{2} + {\left(y - 5\right)}^{2} = 200$

Jan 10, 2016

The circle's equation is ${\left(x - 5\right)}^{2} + {\left(y - 5\right)}^{2} = 200$

#### 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 ${M}_{D E}$
${M}_{D E} \left(\frac{{x}_{D} + {x}_{E}}{2} , \frac{{y}_{D} + {y}_{E}}{2}\right)$ => ${M}_{D} E \left(- 5 , 5\right)$
line 1$\to y = 5$

Equation of Line EF, where ${y}_{E} = {y}_{F} = 15$
$y = 15$
Equation of line 2 perpendicular to EF and passing through midpoint ${M}_{E F}$
${M}_{E F} \left(\frac{{x}_{E} + {x}_{F}}{2} , \frac{{y}_{E} + {y}_{F}}{2}\right)$ => ${M}_{E} F \left(5 , 15\right)$
line 2$\to x = 5$

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

The distance between point C to any of the given points is equal to the circle's radius
$R = {d}_{C D} = \sqrt{{\left(- 5 - 5\right)}^{2} + {\left(- 5 - 5\right)}^{2}} = \sqrt{100 + 100} = \sqrt{200}$

In the formula of the equation of the circle:
${\left(x - {x}_{C}\right)}^{2} + {\left(y - {y}_{C}\right)}^{2} = {R}^{2}$
${\left(x - 5\right)}^{2} + {\left(y - 5\right)}^{2} = 200$