Respuesta :

gmany

[tex] \text{Use:}\\\\(*)\qquad(a+b)^2=a^2+2ab+b^2\\\\(**)\qquad(a-b)^2=a^2-2ab+b^2 [/tex]

[tex] \text{The equation of a center}\\\\(x-h)^2+(y-k)^2=r^2\\\\(h,\ k)-center\\\\r-radius [/tex]

[tex] \text{We have:}\\\\x^2+y^2-6x+10y+34=0\\\\x^2-2\cdot x\cdot3+y^2+2\cdot y\cdot5+34=0\ \ \ \ |+3^2\ |+5^2\\\\\underbrace{x^2-2\cdot x\cdot3+3^2}_{(**)}+\underbrace{y^2+2\cdot y\cdot5+5^2}_{(*)}+34=3^2+5^2\ \ \ \ |-34\\\\(x-3)^2+(y+5)^2=9+25-34\\\\(x-3)^2+(y+5)^2=0 [/tex]

It's the point (3, -5)!