Answer
Circle
Center: $(0,0)$
r=4
See graph
Work Step by Step
We are given the equation:
$x^2+y^2=16$
Put the equation in standard form:
$\dfrac{x^2}{16}+\dfrac{y^2}{16}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Determine $h,k,a,b,c$:
$h=0$
$k=0$
$a^2=16\Rightarrow a=\sqrt{16}=4$
$b^2=16\Rightarrow b=\sqrt {16}=4$
$c^2=a^2-b^2=16-16=0$
As $a=b$, the equation represents a circle of radius $a=b=4$.
The center is:
$(h,k)=(0,0)$
Graph the circle:
