Question

Find the center and the radius of the circle with the equation .

Answer 1

We have been given an equation of circle $x^2+2x+y^2-2y-14=0$. We are asked to find the center and radius of the circle.

First of all, we will write equation of circle is standard form $(x-h)^2+(y-k)^2=r^2$, where center of circle is at point (h,k) and radius is r.

Let us complete the square for both x and y.

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

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

Adding half the square of coefficient of x and y term, we will get:

$(\frac{2}{2})^2=1^2=1$

$(\frac{-2}{2})^2=(-1)^2=1$

$x^2+2x+1+y^2-2y+1=14+1+1$

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

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

Therefore, the center of the circle is $(-1,1)$ and radius is 4 units.