Question

consider a circle whose equation is (x^{2}+y^{2}-2x - 8=0). which statements are true? select three options. the radius of the circle is 3 units. the center of the circle lies on the x - axis. the center of the circle lies on the y - axis. the standard form of the equation is ((x - 1)^{2}+y^{2}=3). the radius of this circle is the same as the radius of the circle whose equation is (x^{2}+y^{2}=9)

Explanation:

Step1: Convert to standard - form

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. Given $x^{2}+y^{2}-2x - 8 = 0$, we complete the square for the $x$ - terms.
$x^{2}-2x=(x - 1)^{2}-1$. So the equation becomes $(x - 1)^{2}-1+y^{2}-8 = 0$, which simplifies to $(x - 1)^{2}+y^{2}=9$.

Step2: Identify center and radius

For the circle $(x - 1)^{2}+y^{2}=9$, the center is $(1,0)$ and the radius $r = 3$ (since $r^{2}=9$, so $r = 3$).

Answer:

1. The radius of the circle is 3 units.
2. The center of the circle lies on the x - axis.
3. The radius of this circle is the same as the radius of the circle whose equation is $x^{2}+y^{2}=9$.