Answered

Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

Mrs. Culland is finding the center of a circle whose equation is [tex]x^2 + y^2 + 6x + 4y - 3 = 0[/tex] by completing the square. Her work is shown:

[tex]
\begin{array}{l}
x^2 + y^2 + 6x + 4y - 3 = 0 \\
x^2 + 6x + y^2 + 4y - 3 = 0 \\
(x^2 + 6x) + (y^2 + 4y) = 3 \\
(x^2 + 6x + 9) + (y^2 + 4y + 4) = 3 + 9 + 4 \\
(x + 3)^2 + (y + 2)^2 = 16
\end{array}
[/tex]

Which completes the work correctly?

A. [tex](x-3)^2 + (y-2)^2 = 4^2[/tex], so the center is [tex](3, 2)[/tex].
B. [tex](x+3)^2 + (y+2)^2 = 4^2[/tex], so the center is [tex](3, 2)[/tex].
C. [tex](x-3)^2 + (y-2)^2 = 4^2[/tex], so the center is [tex](-3, -2)[/tex].
D. [tex](x+3)^2 + (y+2)^2 = 4^2[/tex], so the center is [tex](-3, -2)[/tex].

Sagot :

GinnyAnswer
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]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.