Question

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

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

The general equation of a circle is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center and $r$ is the radius.
For the given equation $x^{2}+y^{2}+2x - 6y-15 = 0$.
Completing the square for x - terms: $x^{2}+2x=(x + 1)^{2}-1$.
Completing the square for y - terms: $y^{2}-6y=(y - 3)^{2}-9$.
So the equation becomes $(x + 1)^{2}-1+(y - 3)^{2}-9-15=0$.

Step2: Rearrange the equation to the standard form of a circle.

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

Answer:

Radius: $5$
Center: $(-1,3)$