Question

find the center and radius of the circle having the given equation. $x^{2}+y^{2}+2x - 6y - 15 = 0$ radius: center: ( )

Explanation:

Step1: Rewrite the equation by completing the square for x - terms.

The x - terms are \(x^{2}+2x\). Completing the square: \(x^{2}+2x=(x + 1)^{2}-1\).

Step2: Rewrite the equation by completing the square for y - terms.

The y - terms are \(y^{2}-6y\). Completing the square: \(y^{2}-6y=(y - 3)^{2}-9\).

Step3: Substitute the completed - square expressions into the original equation.

The original equation \(x^{2}+y^{2}+2x - 6y-15 = 0\) becomes \((x + 1)^{2}-1+(y - 3)^{2}-9-15 = 0\).

Step4: Simplify the equation.

\((x + 1)^{2}+(y - 3)^{2}=1 + 9+15\).
\((x + 1)^{2}+(y - 3)^{2}=25\).

Step5: Identify the center and radius.

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.
For \((x + 1)^{2}+(y - 3)^{2}=25=(x-(-1))^{2}+(y - 3)^{2}=5^{2}\), the center is \((-1,3)\) and the radius \(r = 5\).

Answer:

Radius: 5
Center: (-1,3)