Respuesta :
we are given
[tex] x^2-6x+y^2+10y-2=0 [/tex]
Firstly, we will change it into standard equation of circle
that is
[tex] (x-h)^2 +(y-k)^2 =r^2 [/tex]
where r is radius
(h,k) is center
now, we can change our equation into this form
To change our equation into this form, we need to complete x square and y square
[tex] x^2-6x+y^2+10y-2=0 [/tex]
step-1: Move 2 on right side
[tex] x^2-6x+y^2+10y-2+2=0+2 [/tex]
[tex] x^2-6x+y^2+10y=2 [/tex]
step-2: Complete x square
[tex] x^2-2*3*x+y^2+10y=2 [/tex]
[tex] x^2-2*3*x+3^2+y^2+10y=2+3^2 [/tex]
[tex] (x-3)^2+y^2+10y=2+3^2 [/tex]
step-3: Complete y square
[tex] (x-3)^2+y^2+2*5*y=2+3^2 [/tex]
[tex] (x-3)^2+y^2+2*5*y+5^2=2+3^2+5^2 [/tex]
[tex] (x-3)^2+(y+5)^2=2+3^2+5^2 [/tex]
step-4: Combine right side terms
[tex] (x-3)^2+(y+5)^2=36 [/tex]
[tex] (x-3)^2+(y+5)^2=6^2 [/tex]
now, we can compare with our equation
and we will get
radius =r=6...............Answer
center=(h,k)=(3,-5)........................Answer
