Question

The graph shows a triangle with vertices plotted at (-5, 5), (-5, -3), and (7, -3).

Answer 1

We need to calculate a system of equations using the general formula of the circle

`x^2+y^2+Dx+Ey+F=0`

for the point (-5, 5)

x=-5

y=5

the first equation is

`\begin{gathered} (-5)^2+(5)^2-5D+5E+F=0 \\ 25+25-5D+5E+F=0 \\ 50-5D+5E+F=0 \\ -5D+5E+F=-50 \end{gathered}`

for the point (-5,-3)

x=-5

y=-3

the second equation is

`\begin{gathered} (-5)^2+(-3)^2-5D-3E+F=0 \\ 25+9^{}-5D-3E+F=0 \\ 34^{}-5D-3E+F=0 \\ -5D-3E+F=-34 \end{gathered}`

for the point (7, -3)

x=7

y=-3

the third equation is

`\begin{gathered} (7)^2+(-3)^2+7D-3E+F=0 \\ 49+9+7D-3E+F=0 \\ 58+7D-3E+F=0 \\ 7D-3E+F=-58 \end{gathered}`

Then we solve the system of equations, and we obtain

D=-2

E=-2

F=50

The equation of the circle that passes through these points is

`x^2+y^2-2x-2y-50=0`

for calculate the coordinates of the center of the circle we have

D=-2h

h is the x coordinate of the center of the circle

D=-2

we isolate the h

h=-2/-2=1

E=-2k

k is the y coordinate of the center of the circle

E=-2

we isolate the k

k=-2/-2=1

the center of the circle is (h,k)=(1,1)