Answer:
[tex]x^2+y^2-8x-6y=0[/tex]
Step-by-step explanation:
General Equation Of A Circle
If a,b, and c are real numbers, then the general equation of a circle is
[tex]\displaystyle x^2+y^2+ax+by+c=0[/tex]
There are three unknowns that could eventually be determined if we knew three points of the circle. These points are (1, 7) (8, 6) (7, -1). We only need to replace them in the general equation and solve the resulting system of equations
For the point (1, 7)
[tex](1)^2+(7)^2+a(1)+b(7)+c=0[/tex]
Operating and rearranging
[tex]\displaystyle a+7b+c=-50[/tex]
For the point (8, 6)
[tex]\displaystyle (8)^2+(6)^2+a(8)+b(6)+c=0[/tex]
Operating and rearranging
[tex]\displaystyle 8a+6b+c=-100[/tex]
For the point (7, -1)
[tex]\displaystyle (7)^2+(-1)^2+a(7)+b(-1)+c=0[/tex]
Operating and rearranging
[tex]\displaystyle 7a-b+c=-50[/tex]
We form the system of equations
[tex]\left\{\begin{matrix}a+7b+c=50\\ 8a+6b+c=-100\\ 7a-b+c=-50\end{matrix}\right.[/tex]
We'll eliminate c from the first two equations and then from the last two equations
Multiplying the first one by -1
[tex]\left\{\begin{matrix}-a-7b-c=50\\ 8a+6b+c=-100\end{matrix}\right.[/tex]
Adding them up we have
[tex]\displaystyle 7a-b=-50[/tex]
Adding up with the third equation
[tex]\displaystyle 7a-b+c=-50[/tex]
We get
c=0
Knowing this value, let's return to the original system with c=0
[tex]\left\{\begin{matrix}a+7b=-50\\ 8a+6b=-100\end{matrix}\right.[/tex]
Multiplying the first by -8
[tex]\left\{\begin{matrix}-8a-56b=400\\ 8a+6b=-100\end{matrix}\right.[/tex]
Adding them up
[tex]\displaystyle -50b=300[/tex]
Which gives
[tex]b=-6[/tex]
Finally, isolating a from
[tex]a=-50-7b[/tex]
We get
a=-8
So the general equation of the circle is
[tex]x^2+y^2-8x-6y=0[/tex]
