Question

the equation of a circle is given below. identify the center and the radius. then graph the circle.

$x^{2}+2x + y^{2}-15 = 0$

center: $(-1,0)$

radius: 4

Explanation:

Step1: Complete the square for \(x\) terms

Given equation \(x^{2}+2x + y^{2}-15 = 0\).
For the \(x\) - terms \(x^{2}+2x\), using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), where \(a = x\) and \(2ab=2x\) (so \(b = 1\)).
\(x^{2}+2x=(x + 1)^{2}-1\).
The equation becomes \((x + 1)^{2}-1+y^{2}-15=0\).

Step2: Simplify the equation

Simplify \((x + 1)^{2}-1+y^{2}-15=0\) to \((x + 1)^{2}+y^{2}=16\).
The standard form of a circle's equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.
Comparing \((x + 1)^{2}+y^{2}=16\) (or \((x-(-1))^{2}+(y - 0)^{2}=4^{2}\)) with \((x - h)^{2}+(y - k)^{2}=r^{2}\).

Answer:

Center: \((-1,0)\), Radius: \(4\)