Let's consider the equation of the circle given:
[tex]\[ x^2 + y^2 - 2x - 8 = 0 \][/tex]
We can convert this equation into the standard form of a circle equation [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex] by completing the square.
Step-by-step process:
1. Recognize the general form:
[tex]\[ x^2 + y^2 - 2x - 8 = 0 \][/tex]
2. Complete the square for the [tex]\(x\)[/tex] terms:
- First, isolate the [tex]\(x\)[/tex] terms: [tex]\(x^2 - 2x\)[/tex].
- To complete the square, add and subtract [tex]\((\frac{-2}{2})^2 = 1\)[/tex]:
[tex]\[ x^2 - 2x + 1 - 1 \][/tex]
3. Rewrite the equation with the completed square:
[tex]\[ (x - 1)^2 - 1 + y^2 - 8 = 0 \][/tex]
4. Combine constants on one side:
[tex]\[ (x - 1)^2 + y^2 - 9 = 0 \implies (x - 1)^2 + y^2 = 9 \][/tex]
Now, the equation is in standard form [tex]\((x - 1)^2 + y^2 = 9\)[/tex].
From this standard form, we can derive the properties of the circle:
- The center [tex]\((h, k)\)[/tex] is [tex]\((1, 0)\)[/tex].
- The radius [tex]\(r\)[/tex] is [tex]\(\sqrt{9} = 3\)[/tex].
Now, let's evaluate the given statements:
1. The radius of the circle is 3 units.
- True, because [tex]\(r = \sqrt{9} = 3\)[/tex].
2. The center of the circle lies on the [tex]\(x\)[/tex]-axis.
- True, because the center is at [tex]\((1, 0)\)[/tex] which means [tex]\(k = 0\)[/tex] (on the [tex]\(x\)[/tex]-axis).
3. The center of the circle lies on the [tex]\(y\)[/tex]-axis.
- False, because the center is at [tex]\((1, 0)\)[/tex] which means [tex]\(h \neq 0\)[/tex] (not on the [tex]\(y\)[/tex]-axis).
4. The standard form of the equation is [tex]\((x-1)^2+y^2=3\)[/tex].
- False, as we determined, the correct standard form is [tex]\((x - 1)^2 + y^2 = 9\)[/tex].
5. The radius of this circle is the same as the radius of the circle whose equation is [tex]\(x^2 + y^2 = 9\)[/tex].
- True, because the radius of the circle [tex]\(x^2 + y^2 = 9\)[/tex] is [tex]\(\sqrt{9} = 3\)[/tex].
Therefore, the three correct statements are:
- The radius of the circle is 3 units.
- The center of the circle lies on the [tex]\(x\)[/tex]-axis.
- The radius of this circle is the same as the radius of the circle whose equation is [tex]\(x^2 + y^2 = 9\)[/tex].
