Question

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

Explanation:

Step1: Complete the square for \(x\)

Given \(x^{2}+y^{2}-2x - 8=0\). Group \(x\) - terms: \((x^{2}-2x)+y^{2}=8\).
Using \((a - b)^{2}=a^{2}-2ab + b^{2}\), for \(x^{2}-2x\), \(a = x\), \(2ab=2x\) so \(b = 1\). Then \(x^{2}-2x=(x - 1)^{2}-1\).
The equation becomes \((x - 1)^{2}-1+y^{2}=8\), or \((x - 1)^{2}+y^{2}=9\).

Step2: Identify center and radius

The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.
For \((x - 1)^{2}+y^{2}=9=(x - 1)^{2}+(y - 0)^{2}=3^{2}\), the center is \((1,0)\) and radius \(r = 3\).

• Since the center \((1,0)\) has \(y\) - coordinate \(0\), the center lies on the \(x\) - axis.
• The radius \(r = 3\) units.
• For the circle \(x^{2}+y^{2}=9\), its standard form is \((x - 0)^{2}+(y - 0)^{2}=3^{2}\), radius \(r = 3\). So the radius of \(x^{2}+y^{2}-2x - 8=0\) (which has radius \(3\)) is the same as the radius of \(x^{2}+y^{2}=9\).

Answer:

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