Question 1115398
a)

{{{x^2 + y^2 + 4x -2y +1=0}}} 

{{{(x^2 + 4x) + (y^2 -2y) =-1}}} ....complete squares

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

use the rule: {{{(x+b)^2=a^2+2ab+b^2}}}

since you have first term {{{x^2}}} and middle term {{{4x}}}, means that coefficient {{{a=1}}} and {{{2ab=4}}}=>{{{2b=4}}}=>{{{b=2}}}

and 
since you have first term {{{y^2}}} and middle term {{{-2y}}}, means that coefficient {{{a=1}}} and {{{2ab=-2}}}=>{{{2b=-2}}}=>{{{b=-1}}}

then, you have

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

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

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

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



b) 

{{{x^2 + y^2 - 6x - 8y =0}}} written in the form {{{(x - a)^2 + (y-b))^2= c}}} ( {{{(y+b)^2}}} is wrong, should be {{{(y-b)^2}}})

{{{(x^2  - 6x) + (y^2- 8y) =0}}} 

{{{(x^2  - 6x+b^2)- b^2+ (y^2- 8y+b^2) -b^2=0}}} 

{{{(x^2  - 6x+3^2)- 3^2+ (y^2- 8y+4^2) -4^2=0}}} 

{{{(x  - 3)^2- 9+ (y- 4)^2) -16=0}}} 

{{{(x  - 3)^2+ (y- 4)^2) =9+16}}} 

{{{(x  - 3)^2+ (y- 4)^2) =25}}} 


 so, {{{a=-3}}},{{{b=-4}}} and {{{ c=25}}}