Question 1121380
 to find the center and radius of the following circle, write given equation in this form:

{{{(x-h)^2+(y-k)^2=r^2}}} where {{{h}}} and {{{k}}} are {{{x}}} and {{{y}}} coordinates of the center, and {{{r}}} is the radius


{{{x^2 + y^2 - 12x + 8y - 29 = 0 }}} ......rearrange and group

{{{(x^2  - 12x) + (y^2+ 8y ) = 29 }}}.... complete squares

{{{(x^2  - 12x+b^2)-b^2 + (y^2+ 8y+b^2 ) -b^2= 29 }}}

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

since you have {{{2ab=12}}} and {{{a=1}}}, =>{{{b=6}}} for {{{x}}} part

and {{{2ab=8}}} and {{{a=1}}}, =>{{{b=4}}} for {{{y}}} part

than you have:

{{{(x  - 6)^2-6^2 + (y+ 4)^2  -4^2= 29 }}}

{{{(x  - 6)^2-36 + (y+ 4)^2  -16= 29 }}}

{{{(x  - 6)^2+ (y+ 4)^2  -52= 29 }}}

{{{(x  - 6)^2+ (y+ 4)^2  = 29+52 }}}

{{{(x  - 6)^2+ (y+ 4)^2  = 81 }}}

{{{(x  - 6)^2+ (y+ 4)^2  = 9^2 }}}

now you see that: 

{{{h=6}}} and {{{k=-4}}} and the center is at ({{{6}}}, {{{-4}}})

and  the radius is {{{9}}}