Answer :
To solve this problem, let’s start by rewriting the equation of the given circle in its standard form. The given equation is:
[tex]\[ x^2 + y^2 - 2x - 8 = 0 \][/tex]
We will complete the square to put this equation into the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex].
Completing the square for the [tex]\(x\)[/tex]-terms:
1. Group the [tex]\(x\)[/tex] terms together:
[tex]\[ x^2 - 2x \][/tex]
2. To complete the square, we take the coefficient of [tex]\(x\)[/tex] (which is [tex]\(-2\)[/tex]), halve it to get [tex]\(-1\)[/tex], and then square it to get 1. We add and subtract this squared value inside the equation:
[tex]\[ x^2 - 2x + 1 - 1 \][/tex]
Now we incorporate this into the original equation:
[tex]\[ (x^2 - 2x + 1) - 1 + y^2 - 8 = 0 \][/tex]
This simplifies to:
[tex]\[ (x - 1)^2 - 1 + y^2 - 8 = 0 \][/tex]
3. Simplify further to get the equation in the form of a circle:
[tex]\[ (x - 1)^2 + y^2 - 9 = 0 \][/tex]
[tex]\[ (x - 1)^2 + y^2 = 9 \][/tex]
Now, we have the standard form [tex]\((x - 1)^2 + y^2 = 9\)[/tex].
From the standard form, it is clear that:
- The center of the circle [tex]\((h, k)\)[/tex] is [tex]\((1, 0)\)[/tex].
- The radius [tex]\(r\)[/tex] is the square root of 9, which is 3 units.
Let's analyze the given statements:
1. The radius of the circle is 3 units.
- True. The radius calculated from the standard form is indeed 3 units.
2. The center of the circle lies on the [tex]\(x\)[/tex]-axis.
- True. The center [tex]\((1, 0)\)[/tex] has its [tex]\(y\)[/tex]-coordinate as 0, implying it lies on the [tex]\(x\)[/tex]-axis.
3. The center of the circle lies on the [tex]\(y\)[/tex]-axis.
- False. The center [tex]\((1, 0)\)[/tex] has its [tex]\(x\)[/tex]-coordinate as 1, indicating it does not lie on the [tex]\(y\)[/tex]-axis.
4. The standard form of the equation is [tex]\((x - 1)^2 + y^2 = 3\)[/tex].
- False. The correct standard form obtained 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. The radius of the circle with equation [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].
[tex]\[ x^2 + y^2 - 2x - 8 = 0 \][/tex]
We will complete the square to put this equation into the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex].
Completing the square for the [tex]\(x\)[/tex]-terms:
1. Group the [tex]\(x\)[/tex] terms together:
[tex]\[ x^2 - 2x \][/tex]
2. To complete the square, we take the coefficient of [tex]\(x\)[/tex] (which is [tex]\(-2\)[/tex]), halve it to get [tex]\(-1\)[/tex], and then square it to get 1. We add and subtract this squared value inside the equation:
[tex]\[ x^2 - 2x + 1 - 1 \][/tex]
Now we incorporate this into the original equation:
[tex]\[ (x^2 - 2x + 1) - 1 + y^2 - 8 = 0 \][/tex]
This simplifies to:
[tex]\[ (x - 1)^2 - 1 + y^2 - 8 = 0 \][/tex]
3. Simplify further to get the equation in the form of a circle:
[tex]\[ (x - 1)^2 + y^2 - 9 = 0 \][/tex]
[tex]\[ (x - 1)^2 + y^2 = 9 \][/tex]
Now, we have the standard form [tex]\((x - 1)^2 + y^2 = 9\)[/tex].
From the standard form, it is clear that:
- The center of the circle [tex]\((h, k)\)[/tex] is [tex]\((1, 0)\)[/tex].
- The radius [tex]\(r\)[/tex] is the square root of 9, which is 3 units.
Let's analyze the given statements:
1. The radius of the circle is 3 units.
- True. The radius calculated from the standard form is indeed 3 units.
2. The center of the circle lies on the [tex]\(x\)[/tex]-axis.
- True. The center [tex]\((1, 0)\)[/tex] has its [tex]\(y\)[/tex]-coordinate as 0, implying it lies on the [tex]\(x\)[/tex]-axis.
3. The center of the circle lies on the [tex]\(y\)[/tex]-axis.
- False. The center [tex]\((1, 0)\)[/tex] has its [tex]\(x\)[/tex]-coordinate as 1, indicating it does not lie on the [tex]\(y\)[/tex]-axis.
4. The standard form of the equation is [tex]\((x - 1)^2 + y^2 = 3\)[/tex].
- False. The correct standard form obtained 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. The radius of the circle with equation [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].