Question 1115789

What is the center of the curve

{{{ x^2+y^2-2x-4y-31=0}}}

{{{(x^2-2x)+(y^2-4y)=31}}}....complete square

{{{(x^2-2x+b^2)-b^2+(y^2-4y+b^2) -b^2 =31}}}

recal
{{{(a+b)^2=a^2+2ab+b^2}}}

for {{{x}}} part, you have  {{{a=1}}} and {{{2ab=-2}}}=>{{{2*1*b=-2}}}=>{{{b=-1}}}
for {{{y}}} part, you have  {{{a=1}}} and {{{2ab=-4}}}=>{{{2*1*b=-4}}}=>{{{b=-2}}}

so, you have 
{{{(x^2-2x+(-1)^2)-(-1)^2+(y^2-4y+(-2)^2) -(-2)^2 =31}}}

{{{(x-1)^2-1+(y-2)^2 -(-2)^2 =31}}}

{{{(x-1)^2-1+(y-2)^2 -4 =31}}}

{{{(x-1)^2+(y-2)^2  =31+1+4}}}

{{{(x-1)^2+(y-2)^2  =36}}}

{{{(x-1)^2+(y-2)^2  =6^2}}}

so, you have a circle with  {{{h=1}}}, {{{k=2}}}, and {{{r=6}}}

the center is  at ({{{1}}}, {{{2}}})