To solve this problem, we will first find the standard form of the circle equation. Given a circle of radius r and center (h,k), the standard form of the circle equation would be
[tex](x-h)^2+(y-k)^2=r^2[/tex]In our case, we have h=8 , k=6 and r=10. So the equation for the given circle would be
[tex](x-8)^2+(y-6)^2=10^2=100[/tex]The general form of the circle equation can be obtained from expanding the squares on the left side of the equality sign. Recall that
[tex](a-b)^2=a^2-2a\cdot b+b^2[/tex]So, applying this to the standard equation we get
[tex](x-8)^2=x^2-16x+64[/tex][tex](y-6)^2=y^2-12y+36[/tex]So our equation becomes
[tex]x^2-16x+64+y^2-12y+36=100[/tex]Operating on the left side, we have
[tex]x^2-16x+y^2-12y+100=100[/tex]By subtracting 100 on both sides, we get
[tex]x^2-16x+y^2-12y=0[/tex]which the general form of the equation of the given circle.
Using a graphing tool, we have that the circle's graph would be
