Answer:
The center of circle is [tex](0,0)[/tex]
Step-by-step explanation:
We need to find the center of the circle of the equation [tex]x^{2}+y^{2}=4[/tex]
Since, the general equation of circle is [tex](x-h)^{2}+(y-k)^{2} = r^{2}[/tex]
Where (h,k) is center of circle and r is radius.
Re-write the circle equation is [tex]x^{2}+y^{2}=4[/tex] as,
[tex]x^{2}+y^{2}=2^{2}[/tex]
Compare [tex]x^{2}+y^{2}=2^{2}[/tex] with [tex](x-h)^{2}+(y-k)^{2} = r^{2}[/tex]
so, [tex](x-0)^{2}+(y-0)^{2} = 2^{2}[/tex]
Hence, the center of circle is [tex](0,0)[/tex]
