we know that formula of circle in standard form
[tex](x-h)^2+(y-k)^2=r^2[/tex]
where
(h,k) is center of circle
r is radius of circle
Here , we can see that (x-h)^2 and (y-k)^2 are perfect squares
so, we will change our equations in perfect square
For example:
[tex]x^2+y^2+16x-14y+49=0[/tex]
Firstly, we will change x terms into perfect square
step-1:
we will move constant term on right side
[tex]x^2+y^2+16x-14y+49-49=0-49[/tex]
[tex]x^2+y^2+16x-14y=-49[/tex]
step-2:
We will move all x terms altogether
[tex]x^2+16x+y^2-14y=-49[/tex]
step-3:
Break x terms as 2*a*b
[tex]x^2+2\times 8\times x+y^2-14y=-49[/tex]
step-4:
now, we can use formula
[tex]a^2+2ab+b^2=(a+b)^2[/tex]
we will identify b
so, we get b=8
to make 8^2 , we will add both sides by 8^2
[tex]x^2+2\times 8\times x+8^2+y^2-14y=-49+8^2[/tex]
now, we can write in perfect square form
[tex](x+8)^2+y^2-14y=15[/tex]
step-5:
now, we can break y terms in 2ab form
[tex](x+8)^2+y^2-2\times 7\times y=15[/tex]
step-6:
now, we can use formula
[tex]a^2-2ab+b^2=(a-b)^2[/tex]
we will identify b
so, we get b=7
to make 7^2 , we will add both sides by 7^2
[tex](x+8)^2+y^2-2\times 7\times y+7^2=15+7^2[/tex]
[tex](x+8)^2+(y-7)^2=64[/tex]
[tex](x+8)^2+(y-7)^2=8^2[/tex]
we can see that this is in standard equation of circle form
so, center=(-8,7)
radius=8
