We have been given an equation of circle [tex]x^2+2x+y^2-2y-14=0[/tex]. We are asked to find the center and radius of the circle.
First of all, we will write equation of circle is standard form [tex](x-h)^2+(y-k)^2=r^2[/tex], where center of circle is at point (h,k) and radius is r.
Let us complete the square for both x and y.
[tex]x^2+2x+y^2-2y-14+14=0+14[/tex]
[tex]x^2+2x+y^2-2y=14[/tex]
Adding half the square of coefficient of x and y term, we will get:
[tex](\frac{2}{2})^2=1^2=1[/tex]
[tex](\frac{-2}{2})^2=(-1)^2=1[/tex]
[tex]x^2+2x+1+y^2-2y+1=14+1+1[/tex]
[tex](x+1)^2+(y-1)^2=16[/tex]
[tex](x-(-1))^2+(y-1)^2=4^2[/tex]
Therefore, the center of the circle is [tex](-1,1)[/tex] and radius is 4 units.
