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Sagot :
To find the center and the radius of the circle given by the equation [tex]\( x^2 + y^2 - x - 2y - \frac{11}{4} = 0 \)[/tex], we need to rewrite it in the standard form of a circle’s equation:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
First, let's rewrite and complete the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
1. Completing the Square for [tex]\(x\)[/tex]:
- The terms involving [tex]\(x\)[/tex] in the equation are [tex]\(x^2 - x\)[/tex].
- To complete the square, add and subtract [tex]\((\frac{1}{2})^2\)[/tex]:
[tex]\[ x^2 - x = (x - \frac{1}{2})^2 - \frac{1}{4} \][/tex]
2. Completing the Square for [tex]\(y\)[/tex]:
- The terms involving [tex]\(y\)[/tex] in the equation are [tex]\(y^2 - 2y\)[/tex].
- To complete the square, add and subtract [tex]\(1^2\)[/tex]:
[tex]\[ y^2 - 2y = (y - 1)^2 - 1 \][/tex]
3. Rewriting the Whole Equation:
- Substitute the completed square expressions back into the original equation:
[tex]\[ (x - \frac{1}{2})^2 - \frac{1}{4} + (y - 1)^2 - 1 - \frac{11}{4} = 0 \][/tex]
4. Simplifying:
- Combine the constants on the left side:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{1}{4} - 1 - \frac{11}{4} = 0 \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{1}{4} - 1 - \frac{11}{4} = (x - \frac{1}{2})^2 + (y - 1)^2 - \left(\frac{1}{4} + 1 + \frac{11}{4}\right) = 0 \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{15}{4} = 0 \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = \frac{15}{4} \][/tex]
Now the equation is in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\(h = \frac{1}{2}\)[/tex] and [tex]\(k = 1\)[/tex], and [tex]\(r^2 = \frac{15}{4}\)[/tex].
The center [tex]\((h, k)\)[/tex] of the circle is:
[tex]\[ \left(\frac{1}{2}, 1\right) \][/tex]
The radius [tex]\(r\)[/tex] is:
[tex]\[ r = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2} \approx 1.2247 \][/tex]
Thus, the correct answer with the center and the radius approximately calculated is:
[tex]\[ \boxed{\left(\frac{1}{2}, 1\right), 1.22474487139159 \text{ units}} \][/tex]
The closest answer in the given options is:
[tex]\[ C. \left(\frac{1}{2}, 1\right), 2 \text{ units} \][/tex]
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
First, let's rewrite and complete the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
1. Completing the Square for [tex]\(x\)[/tex]:
- The terms involving [tex]\(x\)[/tex] in the equation are [tex]\(x^2 - x\)[/tex].
- To complete the square, add and subtract [tex]\((\frac{1}{2})^2\)[/tex]:
[tex]\[ x^2 - x = (x - \frac{1}{2})^2 - \frac{1}{4} \][/tex]
2. Completing the Square for [tex]\(y\)[/tex]:
- The terms involving [tex]\(y\)[/tex] in the equation are [tex]\(y^2 - 2y\)[/tex].
- To complete the square, add and subtract [tex]\(1^2\)[/tex]:
[tex]\[ y^2 - 2y = (y - 1)^2 - 1 \][/tex]
3. Rewriting the Whole Equation:
- Substitute the completed square expressions back into the original equation:
[tex]\[ (x - \frac{1}{2})^2 - \frac{1}{4} + (y - 1)^2 - 1 - \frac{11}{4} = 0 \][/tex]
4. Simplifying:
- Combine the constants on the left side:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{1}{4} - 1 - \frac{11}{4} = 0 \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{1}{4} - 1 - \frac{11}{4} = (x - \frac{1}{2})^2 + (y - 1)^2 - \left(\frac{1}{4} + 1 + \frac{11}{4}\right) = 0 \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{15}{4} = 0 \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = \frac{15}{4} \][/tex]
Now the equation is in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\(h = \frac{1}{2}\)[/tex] and [tex]\(k = 1\)[/tex], and [tex]\(r^2 = \frac{15}{4}\)[/tex].
The center [tex]\((h, k)\)[/tex] of the circle is:
[tex]\[ \left(\frac{1}{2}, 1\right) \][/tex]
The radius [tex]\(r\)[/tex] is:
[tex]\[ r = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2} \approx 1.2247 \][/tex]
Thus, the correct answer with the center and the radius approximately calculated is:
[tex]\[ \boxed{\left(\frac{1}{2}, 1\right), 1.22474487139159 \text{ units}} \][/tex]
The closest answer in the given options is:
[tex]\[ C. \left(\frac{1}{2}, 1\right), 2 \text{ units} \][/tex]
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