Question 83616
{{{1 * x ^ 2 + y ^ 2 - 1 6 * x + 1 2 * y  =0}}} Start with the given conic




{{{( x ^ 2 - 1 6 * x ) + ( y ^ 2 + 1 2 * y ) = 0     }}} Group like terms





Now we must complete the individual squares inside the parenthesis:




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 For instance to complete {{{1 ( x ^ 2 - 1 6 * x ) }}}, take half of -16 and square it (ie {{{(-16/2)^2=(-8)^2=64}}} to get 64. Now add 64 inside the parenthesis like this: 

{{{( x ^ 2 - 1 6 * x + 6 4 ) }}}

Since you really added {{{1*64}}} to the entire left side, we must add {{{1*64}}} to the right side also. Now lets complete  {{{ ( y ^ 2 + 1 2 * y )}}}


In order to complete {{{( y ^ 2 + 1 2 * y )}}}, take half of 12 and square it (ie {{{(12/2)^2=(6)^2=36}}} to get 36. Now add 36 inside the parenthesis like this: 

{{{ ( y ^ 2 + 1 2 * y + 3 6 )}}}

Since you really added {{{1*36}}} to the entire left side, we must add {{{1*36}}} to the right side also.


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{{{ ( x ^ 2 - 1 6 * x + 6 4 ) + ( y ^ 2 + 1 2 * y + 3 6 ) = - 0 + 6 4 * 1 + 3 6 * 1 }}} Complete the individual squares by taking half of the 2nd coefficient and squaring it (remember to add to both sides).


{{{ ( x ^ 2 - 1 6 * x + 6 4 ) + ( y ^ 2 + 1 2 * y + 3 6 ) = 1 0 0 }}} Combine like terms on the right side





{{{ ( x - 8 ) ^ 2 +  ( y + 6 ) ^ 2 = 1 0 0 }}} Factor the individual groups on the left side.


So the equation is now in standard form {{{(x-h)^2+(y-k)^2=r^2}}}

where (h,k) is the center and r is the radius


So the equation is a circle with a radius of 10 and a center of (8,-6)


Here's the graph of {{{( x - 8 ) ^ 2 + ( y + 6 ) ^ 2 = 1 0 0 }}}


{{{drawing (500,500,-20, 20, -20, 20,
   circle(8,-6,10),
   circle(8,-6,0.08),
   circle(8,-6,0.10),
   graph( 500, 500, -20, 20, -20, 20,0)
   )
}}}


What is it? Circle
Center: (8,-6)
Radius: 10
Standard Form: {{{( x - 8 ) ^ 2 + ( y + 6 ) ^ 2 = 1 0 0 }}}