Question

Find the center and the radius of the circle with the equation: x^2 +2x + y^2-2y-14 =0

Answer 1

X²+2x+y²-2y=14
adding 1+1 both sides

x²+2x+1+y²-2y+1=14+1+1

(x+1)² +(y-1)² = 4²

x=-1 y= 1 r = 4

hope it helps !!

Answer 2

We need to convert the equation of the circle to standard form only then we can find the radius and the center of the circle.

$x^{2} +2x+ y^{2}-2y-14=0 \\ \\ x^{2} +2x+ y^{2}-2y=14 \\ \\ x^{2} +2(x)(1)+ y^{2}-2(y)(1)=14 \\ \\ x^{2} +2x+1 + (y^{2}-2y+1)=14+1+1 \\ \\ (x+1)^{2}+(y-1)^{2} =16$

From this equation we can write:

Center of the circle is (-1,1) and the radius of the circle is 4.