Answer :
[tex]center\ of\ a\ circle:(a;\ b)\\\\radius:r\\\\(x-a)^2+(y-b)^2=r^2\\\\========================\\\\(h;\ 7);\ r=10\\\\(x-h)^2+(y-7)^2=10^2\\\\========================[/tex]
[tex]The\ circle\ passes\ throught\ (3;-1).\\\\Substitute\ x=3\ and\ y=-1:\\\\(3-h)^2+(-1-7)^2=100\\\\(3-h)^2+(-8)^2=100\\\\(3-h)^2+64=100\ \ \ /-64\\\\(3-h)^2=36\iff3-h=\pm\sqrt{36}\\\\3-h=-6\ or\ 3-h=6\\-h=-6-3\ or\ -h=6-3\\-h=-9\ or\ -h=3\\h=9\ or\ h=-3\\\\Answer:h=9\ or\ h=-3[/tex]
[tex]The\ circle\ passes\ throught\ (3;-1).\\\\Substitute\ x=3\ and\ y=-1:\\\\(3-h)^2+(-1-7)^2=100\\\\(3-h)^2+(-8)^2=100\\\\(3-h)^2+64=100\ \ \ /-64\\\\(3-h)^2=36\iff3-h=\pm\sqrt{36}\\\\3-h=-6\ or\ 3-h=6\\-h=-6-3\ or\ -h=6-3\\-h=-9\ or\ -h=3\\h=9\ or\ h=-3\\\\Answer:h=9\ or\ h=-3[/tex]
[tex]equation\ of\ a\ circle:\\\\(x-h)^2+(y-7)^2=10^2\\\\if\ \ (x;y)=(3;-1),\ then\ \ (3-h)^2+(-1-7)^2=100\\\\(3-h)^2+64=100\\\\ (3-h)^2=36\ \ \ \Leftrightarrow\ \ \ (3-h=6\ \ \ or\ \ \ 3-h=-6)\\\\3-h=6\ \ \ \ \ \Rightarrow\ \ \ h=-3\\3-h=-6\ \ \ \Rightarrow\ \ \ h=9\\\\Ans.\ h=-3\ \ or\ \ h=9[/tex]
