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Sagot :
Alright, let’s break down the given equation of the circle and determine the true statements step-by-step.
1. Given equation:
[tex]\[ x^2 + y^2 - 2x - 8 = 0 \][/tex]
2. Rewriting the equation in standard form:
To convert this equation into the standard form of a circle [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we need to complete the square.
- First, group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms together:
[tex]\[ x^2 - 2x + y^2 = 8 \][/tex]
- Complete the square for [tex]\(x\)[/tex]:
[tex]\[ x^2 - 2x \implies (x - 1)^2 - 1 \][/tex]
Adding and subtracting 1 inside the equation:
[tex]\[ (x - 1)^2 - 1 + y^2 = 8 \][/tex]
- Rearrange it to balance the equation:
[tex]\[ (x - 1)^2 + y^2 = 9 \][/tex]
Now, we have the standard form of the circle’s equation:
[tex]\[ (x - 1)^2 + y^2 = 9 \][/tex]
3. Identify the center and radius:
From the standard form [tex]\((x - 1)^2 + y^2 = 9\)[/tex]:
- The center [tex]\((h, k)\)[/tex] is [tex]\((1, 0)\)[/tex],
- The radius [tex]\(r\)[/tex] is:
[tex]\[ r = \sqrt{9} = 3 \][/tex]
4. Evaluating the given statements:
- The radius of the circle is 3 units.
[tex]\[ \text{True (We found the radius to be 3 units)} \][/tex]
- The center of the circle lies on the [tex]$x$[/tex]-axis.
[tex]\[ \text{False (The center is at } (1, 0) \text{, which is off by one unit along the x-axis)} \][/tex]
- The center of the circle lies on the [tex]$y$[/tex]-axis.
[tex]\[ \text{True (The center is at } (1, 0) \text{, which does not lie on the y-axis)} \][/tex]
- The standard form of the equation is [tex]\((x-1)^2 + y^2 = 3\)[/tex].
[tex]\[ \text{False (The correct standard form is } (x - 1)^2 + y^2 = 9 \text{)} \][/tex]
- 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]\[ \text{True (The radius of the circle } x^2 + y^2 = 9 \text{ is also } \sqrt{9} = 3 \text{ units)} \][/tex]
In summary, the three true statements are:
- The radius of the circle is 3 units.
- The center of the circle lies on the [tex]$y$[/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].
1. Given equation:
[tex]\[ x^2 + y^2 - 2x - 8 = 0 \][/tex]
2. Rewriting the equation in standard form:
To convert this equation into the standard form of a circle [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we need to complete the square.
- First, group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms together:
[tex]\[ x^2 - 2x + y^2 = 8 \][/tex]
- Complete the square for [tex]\(x\)[/tex]:
[tex]\[ x^2 - 2x \implies (x - 1)^2 - 1 \][/tex]
Adding and subtracting 1 inside the equation:
[tex]\[ (x - 1)^2 - 1 + y^2 = 8 \][/tex]
- Rearrange it to balance the equation:
[tex]\[ (x - 1)^2 + y^2 = 9 \][/tex]
Now, we have the standard form of the circle’s equation:
[tex]\[ (x - 1)^2 + y^2 = 9 \][/tex]
3. Identify the center and radius:
From the standard form [tex]\((x - 1)^2 + y^2 = 9\)[/tex]:
- The center [tex]\((h, k)\)[/tex] is [tex]\((1, 0)\)[/tex],
- The radius [tex]\(r\)[/tex] is:
[tex]\[ r = \sqrt{9} = 3 \][/tex]
4. Evaluating the given statements:
- The radius of the circle is 3 units.
[tex]\[ \text{True (We found the radius to be 3 units)} \][/tex]
- The center of the circle lies on the [tex]$x$[/tex]-axis.
[tex]\[ \text{False (The center is at } (1, 0) \text{, which is off by one unit along the x-axis)} \][/tex]
- The center of the circle lies on the [tex]$y$[/tex]-axis.
[tex]\[ \text{True (The center is at } (1, 0) \text{, which does not lie on the y-axis)} \][/tex]
- The standard form of the equation is [tex]\((x-1)^2 + y^2 = 3\)[/tex].
[tex]\[ \text{False (The correct standard form is } (x - 1)^2 + y^2 = 9 \text{)} \][/tex]
- 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]\[ \text{True (The radius of the circle } x^2 + y^2 = 9 \text{ is also } \sqrt{9} = 3 \text{ units)} \][/tex]
In summary, the three true statements are:
- The radius of the circle is 3 units.
- The center of the circle lies on the [tex]$y$[/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].
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