Question

the equation of a circle is given below. identify the center and the radius. then graph the circle.
$x^{2}+y^{2}-2y - 15=0$
center:
radius:

Explanation:

Step1: Rewrite the equation in standard form

The standard - form of a circle equation is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center and $r$ is the radius. Given $x^{2}+y^{2}-2y - 15 = 0$, complete the square for the $y$ - terms.
We have $x^{2}+(y^{2}-2y)=15$. Completing the square for $y^{2}-2y$: add $1$ to both sides of the equation. So $x^{2}+(y^{2}-2y + 1)=15 + 1$.

Step2: Simplify the equation

The left - hand side can be factored as $x^{2}+(y - 1)^{2}=16$.

Answer:

Center: $(0,1)$
Radius: $4$