[tex]The\ equation\ of\ the\ circle:(y-a)^2+(y-b)^2=r^2\\\\(a;\ b)-the\ coordinates\ of\ the\ center\\r-the\ radius\\--------------------------\\The\ distance\ between\ A(x_1;\ y_1)\ and\ B(x_2;\ y_2):\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\---------------------\\The\ distance\ between\ the\ center\ of\ the\ circle\ and\ points\ P\ and\ Q\\is\ equal\ the\ length\ of\ radius.\\\\P(3;\ 5);\ Q(-1;\ 3);\ r=\sqrt{10};\ S(a;\ b)\\\\\sqrt{(3-a)^2+(5-b)^2}=(\sqrt{10})^2\\and\\\sqrt{(-1-a)^2+(3-b)^2}=(\sqrt{10})^2[/tex]
[tex]therefore\\\sqrt{(3-a)^2+(5-b)^2}=\sqrt{(-1-a)^2+(3-b)^2}\\\Downarrow\\(3-a)^2+(5-b)^2=(-1-a)^2+(3-b)^2\\3^2-2\cdot3a+a^2+5^2-2\cdot5b+b^2=1^2+2\cdot1a+a^2+3^2-2\cdot3b+b^2\\9-6a+a^2+25-10b+b^2=1+2a+a^2+9-6b+b^2\\34-6a-10b=10+2a-6b\\24-8a-4b=0\ \ \ |divide\ both\ sides\ by\ 4\\6-2a-b=0\to b=6-2a\\subtitute\ to\ \sqrt{(3-a)^2+(5-b)^2}=(\sqrt{10})^2\\\sqrt{(3-a)^2+(5-(6-2a))^2}=10\\\sqrt{(3-a)^2+(-1+2a)^2}=10\iff(3-a)^2+(2a-1)^2=100\\3^2-2\cdot3a+a^2+(2a)^2-2\cdot2a+1^2=100\\9-6a+a^2+4a^2-4a+1=100[/tex]
[tex]5a^2-10a-90=0\ \ \ |divide\ both\ sides\ by\ 5\\a^2-2a-18=0\\a^2-2a+1-1-18=0\\(a-1)^2=19\to a-1=\sqrt{19}\ or\ a-1=-\sqrt{19}\\a=1+\sqrt{19}\ or\ a=1-\sqrt{19}\\\\b=6-2(1+\sqrt{19})=6-2-2\sqrt{19}=4-2\sqrt{19}\\or\\b=6-2(1-\sqrt{19})=6-2+2\sqrt{19}=4+2\sqrt{19}\\\\Answer:\\(x-1-\sqrt{19})^2+(y-4+2\sqrt{19})^2=(\sqrt{10})^2\\or\\(x-1+\sqrt{19})^2+(y-4-2\sqrt{19})^2=(\sqrt{10})^2[/tex]
