Question

rewrite in standard form: x^2+y^2+4x-6y-12=0

Answer 1

\[\left( x ^{2} +4x\right)+\left( y ^{2}-6y \right)=12\]
complete the squares

Answer 2

add square of half the coefficient of x and similarly for y
\[\left( \frac{ 4 }{ 2} \right)^{2} i.e.,4 and \left( \frac{ -6 }{2 } \right)^{2} i.e.,9\]
\[\left( x ^{2}+4x+4\right)+\left( y ^{2}-6y+9 \right)=12 +4+9\]
\[\left( x+2 \right)^{2}+\left( y-3 \right)^{2}=5^{2}\]
which is a circle with centre (-2,3) and radius=5