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Sagot :
To find the center of the circle given by the equation [tex]\( x^2 + y^2 + 6x + 4y - 3 = 0 \)[/tex], we need to complete the square for both [tex]\( x \)[/tex] and [tex]\( y \)[/tex] variables.
Let's start from the original equation:
[tex]\[ x^2 + y^2 + 6x + 4y - 3 = 0 \][/tex]
Step 1: Reorganize the terms to group the [tex]\( x \)[/tex] terms and [tex]\( y \)[/tex] terms together:
[tex]\[ x^2 + 6x + y^2 + 4y - 3 = 0 \][/tex]
Step 2: Move the constant term to the other side of the equation:
[tex]\[ x^2 + 6x + y^2 + 4y = 3 \][/tex]
Step 3: Complete the square for the [tex]\( x \)[/tex] terms and the [tex]\( y \)[/tex] terms separately.
For the [tex]\( x \)[/tex] terms [tex]\( x^2 + 6x \)[/tex]:
- Take half of the coefficient of [tex]\( x \)[/tex] (which is 6), square it, and add it inside the parentheses:
[tex]\[ x^2 + 6x + 9 \][/tex]
For the [tex]\( y \)[/tex] terms [tex]\( y^2 + 4y \)[/tex]:
- Take half of the coefficient of [tex]\( y \)[/tex] (which is 4), square it, and add it inside the parentheses:
[tex]\[ y^2 + 4y + 4 \][/tex]
However, if we add these terms to the left side, we must also add them to the right side to keep the equation balanced.
Thus:
[tex]\[ (x^2 + 6x + 9) + (y^2 + 4y + 4) = 3 + 9 + 4 \][/tex]
Step 4: Rewrite the completed squares:
[tex]\[ (x + 3)^2 + (y + 2)^2 = 16 \][/tex]
Now, we compare this with the standard form of a circle's equation, [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], which identifies the center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex].
From [tex]\((x + 3)^2 + (y + 2)^2 = 16\)[/tex], it is clear that:
- [tex]\( x + 3 \)[/tex] interprets to [tex]\( x - (-3) \)[/tex], thus [tex]\( h = -3 \)[/tex]
- [tex]\( y + 2 \)[/tex] interprets to [tex]\( y - (-2) \)[/tex], thus [tex]\( k = -2 \)[/tex]
- The right-hand side, 16, is [tex]\( 4^2 \)[/tex] which gives the radius [tex]\(r = 4\)[/tex]
So, the center of the circle is [tex]\((-3, -2)\)[/tex].
Therefore, the correct completed form is [tex]\( (x + 3)^2 + (y + 2)^2 = 4^2 \)[/tex] and the center is [tex]\((-3, -2)\)[/tex].
The correct statement is:
[tex]\[ (x+3)^2+(y+2)^2=4^2\text{, so the center is } (-3,-2). \][/tex]
Let's start from the original equation:
[tex]\[ x^2 + y^2 + 6x + 4y - 3 = 0 \][/tex]
Step 1: Reorganize the terms to group the [tex]\( x \)[/tex] terms and [tex]\( y \)[/tex] terms together:
[tex]\[ x^2 + 6x + y^2 + 4y - 3 = 0 \][/tex]
Step 2: Move the constant term to the other side of the equation:
[tex]\[ x^2 + 6x + y^2 + 4y = 3 \][/tex]
Step 3: Complete the square for the [tex]\( x \)[/tex] terms and the [tex]\( y \)[/tex] terms separately.
For the [tex]\( x \)[/tex] terms [tex]\( x^2 + 6x \)[/tex]:
- Take half of the coefficient of [tex]\( x \)[/tex] (which is 6), square it, and add it inside the parentheses:
[tex]\[ x^2 + 6x + 9 \][/tex]
For the [tex]\( y \)[/tex] terms [tex]\( y^2 + 4y \)[/tex]:
- Take half of the coefficient of [tex]\( y \)[/tex] (which is 4), square it, and add it inside the parentheses:
[tex]\[ y^2 + 4y + 4 \][/tex]
However, if we add these terms to the left side, we must also add them to the right side to keep the equation balanced.
Thus:
[tex]\[ (x^2 + 6x + 9) + (y^2 + 4y + 4) = 3 + 9 + 4 \][/tex]
Step 4: Rewrite the completed squares:
[tex]\[ (x + 3)^2 + (y + 2)^2 = 16 \][/tex]
Now, we compare this with the standard form of a circle's equation, [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], which identifies the center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex].
From [tex]\((x + 3)^2 + (y + 2)^2 = 16\)[/tex], it is clear that:
- [tex]\( x + 3 \)[/tex] interprets to [tex]\( x - (-3) \)[/tex], thus [tex]\( h = -3 \)[/tex]
- [tex]\( y + 2 \)[/tex] interprets to [tex]\( y - (-2) \)[/tex], thus [tex]\( k = -2 \)[/tex]
- The right-hand side, 16, is [tex]\( 4^2 \)[/tex] which gives the radius [tex]\(r = 4\)[/tex]
So, the center of the circle is [tex]\((-3, -2)\)[/tex].
Therefore, the correct completed form is [tex]\( (x + 3)^2 + (y + 2)^2 = 4^2 \)[/tex] and the center is [tex]\((-3, -2)\)[/tex].
The correct statement is:
[tex]\[ (x+3)^2+(y+2)^2=4^2\text{, so the center is } (-3,-2). \][/tex]
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