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.
For \(x^{2}+y^{2}-2x - 8=0\), complete the square for the \(x\) - terms.
\(x^{2}-2x+y^{2}=8\).
\((x - 1)^{2}-1+y^{2}=8\).
\((x - 1)^{2}+y^{2}=9\).

Step2: Analyze the center and radius

From \((x - 1)^{2}+y^{2}=9=(3)^{2}\), the center is \((1,0)\) and the radius \(r = 3\).

• For the center \((1,0)\), since \(y = 0\), the center lies on the \(x\) - axis.
• The standard form is \((x - 1)^{2}+y^{2}=9\), not \((x - 1)^{2}+y^{2}=3\).
• The equation \(x^{2}+y^{2}=9\) also has a radius \(r=\sqrt{9}=3\).

Answer:

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