Hello!
The answer is:
The center of the circle is located on the point (-5,1) and the radius is equal to 3 units.
Why?
To determine the center and the radius of a circle from its equation, we need to look for "h" and "k", being "h" the x-coordinate of the center and "k" the y-coordinate the center, then, calculate the radius.
Since we are given the ordinary equation of the circle, we can find the radius and the center directly.
The ordinary equation is:
[tex](x-h)^{2} +(y-k)^{2} =r^{2}[/tex]
Where,
h is the x-coordinate of the center
y is the y-coordinate of the center
r is the radius.
So, we are given the circle:
[tex](x+5)^{2} +(y-1)^{2} =9[/tex]
Which is also equal to:
[tex](x-(-5))^{2} +(y-(1))^{2} =9[/tex]
Where,
[tex]h=-5\\k=1[/tex]
[tex]r^{2}=9\\\sqrt{r^{2}}=\sqrt{9} \\r=3units[/tex]
Hence, the center of the circle is located on the point (-5,1) and the radius is equal to 3 units.
Have a nice day!
