The center of the circle is (2,-3) while the radius of the circle is 2 units
Here, we want to determine the center and the radius of the given circle
Generally, we have the equation of a circle as follows;
[tex](x-a)^2+(y-b)^2=r^2[/tex]Where (a,b) represents the center of the circle and r is the radius of the circle
We have this as follows by dividing the coefficients of x and y by 2
[tex]\begin{gathered} (x+2)^2+(y-3)^2-4\text{ = 0} \\ \text{where (x+2)}^2=x^2+4x\text{ + 4} \\ (y-3)^2=y^2-6y+9 \\ By\text{ subtracting 4 from the sum 13, we have -4} \\ so\text{ we have;} \\ (x+2)^2+(y-3)^2\text{ = 4} \\ (x+2)^2+(y-3)^2=2^2 \end{gathered}[/tex]The center of the circle is (2,-3) while the radius of the circle is 2 units
