Question

Need help with 12(i). Tks!

Answer 1

`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=√((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=√(10);\ S(a;\ b)\\\\√((3-a)^2+(5-b)^2)=(√(10))^2\\and\\√((-1-a)^2+(3-b)^2)=(√(10))^2`

`therefore\\√((3-a)^2+(5-b)^2)=√((-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\ √((3-a)^2+(5-b)^2)=(√(10))^2\\√((3-a)^2+(5-(6-2a))^2)=10\\√((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`

`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=√(19)\ or\ a-1=-√(19)\\a=1+√(19)\ or\ a=1-√(19)\\\\b=6-2(1+√(19))=6-2-2√(19)=4-2√(19)\\or\\b=6-2(1-√(19))=6-2+2√(19)=4+2√(19)\\\\Answer:\\(x-1-√(19))^2+(y-4+2√(19))^2=(√(10))^2\\or\\(x-1+√(19))^2+(y-4-2√(19))^2=(√(10))^2`