Answer
$(a)$
The Center is at $O(0,0)$
$r=5$
$(b)$
The center is at $A(2,-1)$
The radius is $r=3$
$(c)$
The center is at $B(-3,1)$
The radius is $r=2$
Work Step by Step
In general, an equation of a circle has a form: $(x – a)^2 + (y – b)^2 = r^2$
Where $a$ and $b$ are coordinates of the center of the circle.
$r$ is radius
---
$(a)$ $x^2+y^2=25$
To make it look more like a general form:
$x^2+y^2=5^2$
As we can clearly notice, we have no $a$ and $b$ values, so they are $0$, which means that the circle has center at the origin $O(0,0)$
And the radius $r=5$
See the graph $(a)$ on the image above.
$(b)$
$(x – 2)^2 + (y+1)^2 = 9$
As we see:
$a=2$
$b=-1$
So, the center is at $A(2,-1)$
The radius is $r=3$
See the graph $(b)$ on the image above.
$(c)$
$x^2+6x+y^2-2y+6=0$
To find the coordinates of center and radius, we have to simplify the equation and make it look like a general equation.
$(x^2+6x+9)+(y^2-2y+1)=+9+1-6$
$(x+3)^2+(y-1)^2=4$
As we see, the center is at $B(-3, 1)$
$r=2$
See the graph $(c)$ on the image above.
