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Sagot :
Let's start by rewriting the given equation of the circle in standard form. The given equation is:
[tex]\[ x^2 + y^2 - 2x - 8 = 0 \][/tex]
### Step-by-Step Solution:
1. Arrange the equation to separate the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ x^2 - 2x + y^2 - 8 = 0 \][/tex]
2. Complete the square for the [tex]\(x\)[/tex]-terms:
To complete the square, add and subtract [tex]\((\frac{-2}{2})^2 = 1\)[/tex]:
[tex]\[ x^2 - 2x + 1 - 1 + y^2 - 8 = 0 \][/tex]
The expression [tex]\(x^2 - 2x + 1\)[/tex] can be rewritten as [tex]\((x - 1)^2\)[/tex]:
[tex]\[ (x - 1)^2 - 1 + y^2 - 8 = 0 \][/tex]
3. Combine the constants on the right side:
[tex]\[ (x - 1)^2 + y^2 - 9 = 0 \][/tex]
[tex]\[ (x - 1)^2 + y^2 = 9 \][/tex]
This is the standard form of the circle's equation. From this, we can identify the center and the radius of the circle:
- The standard form of a circle's equation is [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center and [tex]\(r\)[/tex] is the radius.
Comparing [tex]\((x - 1)^2 + y^2 = 9\)[/tex] with [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we get:
- Center [tex]\((h, k)\)[/tex] is [tex]\((1, 0)\)[/tex]
- Radius [tex]\(r\)[/tex] is [tex]\(\sqrt{9} = 3\)[/tex]
### Evaluate the Statements:
1. The radius of the circle is 3 units.
This statement is true because we found that the radius [tex]\(r\)[/tex] is 3 units.
2. The center of the circle lies on the [tex]\(x\)[/tex]-axis.
This statement is true because the center [tex]\((1, 0)\)[/tex] has its [tex]\(y\)[/tex]-coordinate as 0, meaning it lies on the [tex]\(x\)[/tex]-axis.
3. The center of the circle lies on the [tex]\(y\)[/tex]-axis.
This statement is false because the center [tex]\((1, 0)\)[/tex] has its [tex]\(x\)[/tex]-coordinate as 1, not 0, meaning 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].
This statement is false because the correct standard form of the equation is [tex]\((x-1)^2 + y^2 = 9\)[/tex], not [tex]\((x-1)^2 + y^2 = 3\)[/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].
The circle [tex]\(x^2 + y^2 = 9\)[/tex] has a radius of [tex]\(\sqrt{9} = 3\)[/tex]. Since the radius of our circle is also 3 units, this statement is true.
### Final Answer:
The three correct statements are:
1. The radius of the circle is 3 units.
2. The center of the circle lies on the [tex]\(x\)[/tex]-axis.
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].
[tex]\[ x^2 + y^2 - 2x - 8 = 0 \][/tex]
### Step-by-Step Solution:
1. Arrange the equation to separate the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ x^2 - 2x + y^2 - 8 = 0 \][/tex]
2. Complete the square for the [tex]\(x\)[/tex]-terms:
To complete the square, add and subtract [tex]\((\frac{-2}{2})^2 = 1\)[/tex]:
[tex]\[ x^2 - 2x + 1 - 1 + y^2 - 8 = 0 \][/tex]
The expression [tex]\(x^2 - 2x + 1\)[/tex] can be rewritten as [tex]\((x - 1)^2\)[/tex]:
[tex]\[ (x - 1)^2 - 1 + y^2 - 8 = 0 \][/tex]
3. Combine the constants on the right side:
[tex]\[ (x - 1)^2 + y^2 - 9 = 0 \][/tex]
[tex]\[ (x - 1)^2 + y^2 = 9 \][/tex]
This is the standard form of the circle's equation. From this, we can identify the center and the radius of the circle:
- The standard form of a circle's equation is [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center and [tex]\(r\)[/tex] is the radius.
Comparing [tex]\((x - 1)^2 + y^2 = 9\)[/tex] with [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we get:
- Center [tex]\((h, k)\)[/tex] is [tex]\((1, 0)\)[/tex]
- Radius [tex]\(r\)[/tex] is [tex]\(\sqrt{9} = 3\)[/tex]
### Evaluate the Statements:
1. The radius of the circle is 3 units.
This statement is true because we found that the radius [tex]\(r\)[/tex] is 3 units.
2. The center of the circle lies on the [tex]\(x\)[/tex]-axis.
This statement is true because the center [tex]\((1, 0)\)[/tex] has its [tex]\(y\)[/tex]-coordinate as 0, meaning it lies on the [tex]\(x\)[/tex]-axis.
3. The center of the circle lies on the [tex]\(y\)[/tex]-axis.
This statement is false because the center [tex]\((1, 0)\)[/tex] has its [tex]\(x\)[/tex]-coordinate as 1, not 0, meaning 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].
This statement is false because the correct standard form of the equation is [tex]\((x-1)^2 + y^2 = 9\)[/tex], not [tex]\((x-1)^2 + y^2 = 3\)[/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].
The circle [tex]\(x^2 + y^2 = 9\)[/tex] has a radius of [tex]\(\sqrt{9} = 3\)[/tex]. Since the radius of our circle is also 3 units, this statement is true.
### Final Answer:
The three correct statements are:
1. The radius of the circle is 3 units.
2. The center of the circle lies on the [tex]\(x\)[/tex]-axis.
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].
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