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Sagot :
To find the radius of the circle from the given equation:
[tex]\[ x^2 + y^2 + 8x - 6y + 21 = 0 \][/tex]
we need to rewrite the equation in the standard form of a circle's equation, [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex]. We achieve this by completing the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
1. Start by grouping the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms together and move the constant to the other side:
[tex]\[ (x^2 + 8x) + (y^2 - 6y) = -21 \][/tex]
2. Complete the square for the [tex]\(x\)[/tex] terms. To do this, take half of the coefficient of [tex]\(x\)[/tex], square it, and then add and subtract this value inside the parentheses:
The coefficient of [tex]\(x\)[/tex] is 8. Half of this is 4, and squaring it gives [tex]\(4^2 = 16\)[/tex]:
[tex]\[ x^2 + 8x = (x + 4)^2 - 16 \][/tex]
3. Similarly, complete the square for the [tex]\(y\)[/tex] terms. Take half of the coefficient of [tex]\(y\)[/tex], square it, and then add and subtract this value inside the parentheses:
The coefficient of [tex]\(y\)[/tex] is -6. Half of this is -3, and squaring it gives [tex]\((-3)^2 = 9\)[/tex]:
[tex]\[ y^2 - 6y = (y - 3)^2 - 9 \][/tex]
4. Substitute these completed squares back into the equation:
[tex]\[ (x + 4)^2 - 16 + (y - 3)^2 - 9 = -21 \][/tex]
5. Combine the constants on the right side of the equation:
[tex]\[ (x + 4)^2 - 16 + (y - 3)^2 - 9 = -21 \][/tex]
[tex]\[ (x + 4)^2 + (y - 3)^2 - 25 = -21 \][/tex]
[tex]\[ (x + 4)^2 + (y - 3)^2 = -21 + 25 \][/tex]
[tex]\[ (x + 4)^2 + (y - 3)^2 = 4 \][/tex]
6. The standard form of a circle's equation is [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]. Here, [tex]\((h, k)\)[/tex] is the center and [tex]\(r\)[/tex] is the radius. Comparing [tex]\((x + 4)^2 + (y - 3)^2 = 4\)[/tex] with the standard form, we can see:
[tex]\[ r^2 = 4 \][/tex]
[tex]\[ r = \sqrt{4} = 2 \][/tex]
Thus, the radius of the circle is 2 units.
The correct answer is:
[tex]\[ \boxed{2 \text{ units}} \][/tex]
[tex]\[ x^2 + y^2 + 8x - 6y + 21 = 0 \][/tex]
we need to rewrite the equation in the standard form of a circle's equation, [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex]. We achieve this by completing the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
1. Start by grouping the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms together and move the constant to the other side:
[tex]\[ (x^2 + 8x) + (y^2 - 6y) = -21 \][/tex]
2. Complete the square for the [tex]\(x\)[/tex] terms. To do this, take half of the coefficient of [tex]\(x\)[/tex], square it, and then add and subtract this value inside the parentheses:
The coefficient of [tex]\(x\)[/tex] is 8. Half of this is 4, and squaring it gives [tex]\(4^2 = 16\)[/tex]:
[tex]\[ x^2 + 8x = (x + 4)^2 - 16 \][/tex]
3. Similarly, complete the square for the [tex]\(y\)[/tex] terms. Take half of the coefficient of [tex]\(y\)[/tex], square it, and then add and subtract this value inside the parentheses:
The coefficient of [tex]\(y\)[/tex] is -6. Half of this is -3, and squaring it gives [tex]\((-3)^2 = 9\)[/tex]:
[tex]\[ y^2 - 6y = (y - 3)^2 - 9 \][/tex]
4. Substitute these completed squares back into the equation:
[tex]\[ (x + 4)^2 - 16 + (y - 3)^2 - 9 = -21 \][/tex]
5. Combine the constants on the right side of the equation:
[tex]\[ (x + 4)^2 - 16 + (y - 3)^2 - 9 = -21 \][/tex]
[tex]\[ (x + 4)^2 + (y - 3)^2 - 25 = -21 \][/tex]
[tex]\[ (x + 4)^2 + (y - 3)^2 = -21 + 25 \][/tex]
[tex]\[ (x + 4)^2 + (y - 3)^2 = 4 \][/tex]
6. The standard form of a circle's equation is [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]. Here, [tex]\((h, k)\)[/tex] is the center and [tex]\(r\)[/tex] is the radius. Comparing [tex]\((x + 4)^2 + (y - 3)^2 = 4\)[/tex] with the standard form, we can see:
[tex]\[ r^2 = 4 \][/tex]
[tex]\[ r = \sqrt{4} = 2 \][/tex]
Thus, the radius of the circle is 2 units.
The correct answer is:
[tex]\[ \boxed{2 \text{ units}} \][/tex]
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