14 (a) The Equation Of A Circle Is Given By X 2 − 6 X + Y 2 + 4 Y − 3 = 0 X^2 - 6x + Y^2 + 4y - 3 = 0 X 2 − 6 X + Y 2 + 4 Y − 3 = 0 . Determine The Coordinates Of The Center And The Radius Of The Circle.

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Introduction

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The equation of a circle in the Cartesian plane is given by $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ represents the coordinates of the center of the circle and $r$ is the radius. In this problem, we are given the equation of a circle in the form $x^2 - 6x + y^2 + 4y - 3 = 0$. Our objective is to determine the coordinates of the center and the radius of the circle.

Completing the Square

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To find the center and radius of the circle, we need to rewrite the given equation in the standard form of the equation of a circle. We can do this by completing the square for both the $x$ and $y$ terms.

Completing the Square for the x-Term

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The given equation is $x^2 - 6x + y^2 + 4y - 3 = 0$. To complete the square for the $x$-term, we need to add and subtract the square of half the coefficient of the $x$-term. The coefficient of the $x$-term is $-6$, so half of this is $-3$, and the square of this is $(-3)^2 = 9$. We add and subtract $9$ inside the parentheses:

(x^2 - 6x + 9) - 9 + y^2 + 4y - 3 = 0

Simplifying the x-Term

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Now, we can simplify the $x$-term by recognizing that it is a perfect square:

(x - 3)^2 - 9 + y^2 + 4y - 3 = 0

Completing the Square for the y-Term

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To complete the square for the $y$-term, we need to add and subtract the square of half the coefficient of the $y$-term. The coefficient of the $y$-term is $4$, so half of this is $2$, and the square of this is $2^2 = 4$. We add and subtract $4$ inside the parentheses:

(x - 3)^2 - 9 + (y^2 + 4y + 4) - 4 - 3 = 0

Simplifying the y-Term

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Now, we can simplify the $y$-term by recognizing that it is a perfect square:

(x - 3)^2 - 9 + (y + 2)^2 - 4 - 3 = 0

Combining the Terms

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Now, we can combine the terms:

(x - 3)^2 + (y + 2)^2 - 16 = 0

Adding 16 to Both Sides

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To isolate the squared terms, we add $16$ to both sides of the equation:

(x - 3)^2 + (y + 2)^2 = 16

Finding the Center and Radius

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Now that we have the equation in the standard form, we can easily identify the center and radius of the circle. The center of the circle is given by the coordinates $(h, k)$, which in this case is $(3, -2)$. The radius of the circle is given by the square root of the constant term on the right-hand side of the equation, which in this case is $\sqrt{16} = 4$.

Conclusion

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In this problem, we were given the equation of a circle in the form $x^2 - 6x + y^2 + 4y - 3 = 0$. We completed the square for both the $x$ and $y$ terms to rewrite the equation in the standard form of the equation of a circle. We then identified the center and radius of the circle as $(3, -2)$ and $4$, respectively.

Final Answer

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The final answer is: $\boxed{(3, -2), 4}$

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Introduction

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In the previous article, we discussed how to find the center and radius of a circle given its equation in the form $x^2 - 6x + y^2 + 4y - 3 = 0$. We completed the square for both the $x$ and $y$ terms to rewrite the equation in the standard form of the equation of a circle. In this article, we will answer some frequently asked questions related to the equation of a circle.

Q&A

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Q: What is the standard form of the equation of a circle?

A: The standard form of the equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ represents the coordinates of the center of the circle and $r$ is the radius.

Q: How do I complete the square for the x-term?

A: To complete the square for the $x$-term, you need to add and subtract the square of half the coefficient of the $x$-term. The coefficient of the $x$-term is $-6$, so half of this is $-3$, and the square of this is $(-3)^2 = 9$. You add and subtract $9$ inside the parentheses.

Q: How do I complete the square for the y-term?

A: To complete the square for the $y$-term, you need to add and subtract the square of half the coefficient of the $y$-term. The coefficient of the $y$-term is $4$, so half of this is $2$, and the square of this is $2^2 = 4$. You add and subtract $4$ inside the parentheses.

Q: What is the center of the circle?

A: The center of the circle is given by the coordinates $(h, k)$, which can be found by completing the square for both the $x$ and $y$ terms.

Q: What is the radius of the circle?

A: The radius of the circle is given by the square root of the constant term on the right-hand side of the equation.

Q: How do I find the equation of a circle given its center and radius?

A: To find the equation of a circle given its center and radius, you can use the standard form of the equation of a circle: $(x - h)^2 + (y - k)^2 = r^2$. Simply plug in the values of the center $(h, k)$ and the radius $r$.

Q: What are some common mistakes to avoid when completing the square?

A: Some common mistakes to avoid when completing the square include:

• Not adding and subtracting the same value inside the parentheses
• Not using the correct value to complete the square
• Not simplifying the equation after completing the square

Conclusion

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In this article, we answered some frequently asked questions related to the equation of a circle. We discussed how to complete the square for both the $x$ and $y$ terms, how to find the center and radius of the circle, and how to find the equation of a circle given its center and radius. We also highlighted some common mistakes to avoid when completing the square.

Final Answer

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The final answer is: \boxedNo final answer, this is a Q&A article.}