What Is The Radius Of A Circle Whose Equation Is $x^2 + Y^2 + 8x - 6y + 21 = 0$?A. 2 Units B. 3 Units C. 4 Units D. 5 Units

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Introduction

In mathematics, the equation of a circle is a fundamental concept that is used to describe the shape and size of a circle. The general equation of a circle is given by $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius of the circle. In this article, we will discuss how to find the radius of a circle whose equation is given as $x^2 + y^2 + 8x - 6y + 21 = 0$.

Understanding the Equation of a Circle

The equation of a circle can be written in the standard form $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius of the circle. To find the radius of a circle, we need to rewrite the equation in the standard form. The given equation is $x^2 + y^2 + 8x - 6y + 21 = 0$. To rewrite this equation in the standard form, we need to complete the square for both the $x$ and $y$ terms.

Completing the Square for the $x$ and $y$ Terms

To complete the square for the $x$ term, we need to add and subtract $(8/2)^2 = 16$ inside the equation. Similarly, to complete the square for the $y$ term, we need to add and subtract $(-6/2)^2 = 9$ inside the equation. The equation becomes:

$x^2 + 8x + 16 + y^2 - 6y + 9 + 21 - 16 - 9 = 0$

Simplifying the equation, we get:

$(x + 4)^2 + (y - 3)^2 = 4$

Finding the Radius of the Circle

Now that we have rewritten the equation in the standard form, we can easily find the radius of the circle. The radius of the circle is the square root of the constant term on the right-hand side of the equation. In this case, the constant term is $4$, so the radius of the circle is $\sqrt{4} = 2$ units.

Conclusion

In this article, we discussed how to find the radius of a circle whose equation is given as $x^2 + y^2 + 8x - 6y + 21 = 0$. We completed the square for both the $x$ and $y$ terms to rewrite the equation in the standard form, and then found the radius of the circle by taking the square root of the constant term on the right-hand side of the equation. The radius of the circle is $\boxed{2}$ units.

Frequently Asked Questions

• What is the equation of a circle? The equation of a circle is given by $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius of the circle.
• How do I find the radius of a circle? To find the radius of a circle, you need to rewrite the equation in the standard form by completing the square for both the $x$ and $y$ terms, and then take the square root of the constant term on the right-hand side of the equation.
• What is the radius of the circle whose equation is $x^2 + y^2 + 8x - 6y + 21 = 0$? The radius of the circle is $\boxed{2}$ units.

Step-by-Step Solution

1. Rewrite the equation in the standard form by completing the square for both the $x$ and $y$ terms.
2. Simplify the equation to get $(x + 4)^2 + (y - 3)^2 = 4$.
3. Find the radius of the circle by taking the square root of the constant term on the right-hand side of the equation.
4. The radius of the circle is $\boxed{2}$ units.

Example Problems

• Find the radius of the circle whose equation is $x^2 + y^2 - 6x + 4y + 12 = 0$. To find the radius of the circle, we need to complete the square for both the $x$ and $y$ terms. The equation becomes $(x - 3)^2 + (y + 2)^2 = 1$. The radius of the circle is $\sqrt{1} = 1$ unit.
• Find the radius of the circle whose equation is $x^2 + y^2 + 2x - 4y + 8 = 0$. To find the radius of the circle, we need to complete the square for both the $x$ and $y$ terms. The equation becomes $(x + 1)^2 + (y - 2)^2 = 1$. The radius of the circle is $\sqrt{1} = 1$ unit.

Tips and Tricks

• To find the radius of a circle, you need to rewrite the equation in the standard form by completing the square for both the $x$ and $y$ terms.
• The radius of the circle is the square root of the constant term on the right-hand side of the equation.
• You can use the equation of a circle to find the radius of a circle in a variety of situations, such as in geometry and trigonometry problems.
  

Introduction

In our previous article, we discussed how to find the radius of a circle whose equation is given as $x^2 + y^2 + 8x - 6y + 21 = 0$. We completed the square for both the $x$ and $y$ terms to rewrite the equation in the standard form, and then found the radius of the circle by taking the square root of the constant term on the right-hand side of the equation. In this article, we will answer some frequently asked questions (FAQs) about the radius of a circle.

Q&A

Q: What is the equation of a circle?

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

Q: How do I find the radius of a circle?

A: To find the radius of a circle, you need to rewrite the equation in the standard form by completing the square for both the $x$ and $y$ terms, and then take the square root of the constant term on the right-hand side of the equation.

Q: What is the radius of the circle whose equation is $x^2 + y^2 + 8x - 6y + 21 = 0$?

A: The radius of the circle is $\boxed{2}$ units.

Q: How do I complete the square for the $x$ and $y$ terms?

A: To complete the square for the $x$ term, you need to add and subtract $(8/2)^2 = 16$ inside the equation. Similarly, to complete the square for the $y$ term, you need to add and subtract $(-6/2)^2 = 9$ inside the equation.

Q: What is the significance of the constant term on the right-hand side of the equation?

A: The constant term on the right-hand side of the equation represents the square of the radius of the circle.

Q: Can I use the equation of a circle to find the radius of a circle in a variety of situations?

A: Yes, you can use the equation of a circle to find the radius of a circle in a variety of situations, such as in geometry and trigonometry problems.

Q: How do I determine the center of the circle?

A: To determine the center of the circle, you need to rewrite the equation in the standard form by completing the square for both the $x$ and $y$ terms. The center of the circle is given by $(h, k)$, where $h$ and $k$ are the values that are added and subtracted inside the equation.

Q: Can I use the equation of a circle to find the area of the circle?

A: Yes, you can use the equation of a circle to find the area of the circle. The area of the circle is given by $\pi r^2$, where $r$ is the radius of the circle.

Q: How do I find the circumference of the circle?

A: To find the circumference of the circle, you need to use the formula $C = 2\pi r$, where $r$ is the radius of the circle.

Example Problems

• Find the radius of the circle whose equation is $x^2 + y^2 - 6x + 4y + 12 = 0$. To find the radius of the circle, we need to complete the square for both the $x$ and $y$ terms. The equation becomes $(x - 3)^2 + (y + 2)^2 = 1$. The radius of the circle is $\sqrt{1} = 1$ unit.
• Find the radius of the circle whose equation is $x^2 + y^2 + 2x - 4y + 8 = 0$. To find the radius of the circle, we need to complete the square for both the $x$ and $y$ terms. The equation becomes $(x + 1)^2 + (y - 2)^2 = 1$. The radius of the circle is $\sqrt{1} = 1$ unit.

Tips and Tricks

• To find the radius of a circle, you need to rewrite the equation in the standard form by completing the square for both the $x$ and $y$ terms.
• The radius of the circle is the square root of the constant term on the right-hand side of the equation.
• You can use the equation of a circle to find the radius of a circle in a variety of situations, such as in geometry and trigonometry problems.

Common Mistakes to Avoid

• Not completing the square for both the $x$ and $y$ terms.
• Not taking the square root of the constant term on the right-hand side of the equation.
• Not using the correct formula to find the area and circumference of the circle.

Conclusion

In this article, we answered some frequently asked questions (FAQs) about the radius of a circle. We discussed how to find the radius of a circle by completing the square for both the $x$ and $y$ terms, and then taking the square root of the constant term on the right-hand side of the equation. We also provided some example problems and tips and tricks to help you understand the concept better.