# A circle has the equation x^2+y^2+2x-2y-14=0, determine the center and radius.

### 2 Answers | Add Yours

The equation of the circle is x^2+y^2+2x-2y-14=0. Write this in the standard form of a circle with center (h, k) and radius r: (x - h)^2 + (y - k)^2 = r^2

x^2+y^2+2x-2y-14=0

=> x^2 + 2x + 1 + y^2 - 2y + 1 = 16

=> (x + 1)^2 + (y - 1)^2 = 4^2

**The center of the circle is (-1, 1) and the radius is 4. **

## A circle has the equation x^2+y^2+2x-2y-14=0, determine the center and radius.

Let the centre of the circle (h, k) and radious is "r".The general equation of the circle

(x-h)^2 +(y-k)^2=r^2. you have to write the given equation in the general equation of the circle form

## x^2+y^2+2x-2y-14=0

(x^2+2.x.1+1^2-1^2)+(y^2-2.y.1+1^2-1^1)=14

(x+1)^2-1+(y-1)^2-1=14

(x+1)^2 + (y-1)^2=14+2

(x+1)^2 + (y-1)^2=16

(x+1)^2 + (y-1)^2=4^2

there fore compare with general equation centre(-1,1) and radius is 4

**Sources:**