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

What is the Radius of a Circle Whose Equation is $x^2 + y^2 + 8x - 6y + 21 = 0$?

Understanding the Equation of a Circle

The equation of a circle in standard form 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. To find the radius of a circle from its equation, we need to rewrite the equation in standard form.

Rewriting the Equation in Standard Form

The given equation is $x^2 + y^2 + 8x - 6y + 21 = 0$. To rewrite this equation in standard form, we need to complete the square for both the $x$ and $y$ terms.

Completing the Square for the $x$ Terms

To complete the square for the $x$ terms, we need to add $(8/2)^2 = 16$ to both sides of the equation.

x^2 + 8x + 16 + y^2 - 6y + 21 = 16

Completing the Square for the $y$ Terms

To complete the square for the $y$ terms, we need to add $(-6/2)^2 = 9$ to both sides of the equation.

x^2 + 8x + 16 + y^2 - 6y + 9 = 16 + 9

Simplifying the Equation

Now, we can simplify the equation by combining like terms.

(x + 4)^2 + (y - 3)^2 = 25

Identifying the Center and Radius

Comparing this equation with the standard form of the equation of a circle, we can see that the center of the circle is $(-4, 3)$ and the radius is $\sqrt{25} = 5$ units.

Conclusion

Therefore, the radius of the circle whose equation is $x^2 + y^2 + 8x - 6y + 21 = 0$ is $\boxed{5}$ units.

Key Takeaways

• To find the radius of a circle from its equation, we need to rewrite the equation in standard form.
• Completing the square for both the $x$ and $y$ terms is a crucial step in rewriting the equation in standard form.
• The center of the circle is given by the values of $h$ and $k$ in the standard form of the equation.
• The radius of the circle is given by the value of $r$ in the standard form of the equation.

Frequently Asked Questions

• What is the equation of a circle in standard form?
• How do we complete the square for the $x$ and $y$ terms?
• What is the center of the circle in the given equation?
• What is the radius of the circle in the given equation?

References

• [1] "Equation of a Circle" by Math Open Reference
• [2] "Completing the Square" by Khan Academy
• [3] "Standard Form of the Equation of a Circle" by Purplemath
  What is the Radius of a Circle Whose Equation is $x^2 + y^2 + 8x - 6y + 21 = 0$?

Understanding the Equation of a Circle

The equation of a circle in standard form 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. To find the radius of a circle from its equation, we need to rewrite the equation in standard form.

Rewriting the Equation in Standard Form

The given equation is $x^2 + y^2 + 8x - 6y + 21 = 0$. To rewrite this equation in standard form, we need to complete the square for both the $x$ and $y$ terms.

Completing the Square for the $x$ Terms

To complete the square for the $x$ terms, we need to add $(8/2)^2 = 16$ to both sides of the equation.

x^2 + 8x + 16 + y^2 - 6y + 21 = 16

Completing the Square for the $y$ Terms

To complete the square for the $y$ terms, we need to add $(-6/2)^2 = 9$ to both sides of the equation.

x^2 + 8x + 16 + y^2 - 6y + 9 = 16 + 9

Simplifying the Equation

Now, we can simplify the equation by combining like terms.

(x + 4)^2 + (y - 3)^2 = 25

Identifying the Center and Radius

Comparing this equation with the standard form of the equation of a circle, we can see that the center of the circle is $(-4, 3)$ and the radius is $\sqrt{25} = 5$ units.

Conclusion

Therefore, the radius of the circle whose equation is $x^2 + y^2 + 8x - 6y + 21 = 0$ is $\boxed{5}$ units.

Key Takeaways

• To find the radius of a circle from its equation, we need to rewrite the equation in standard form.
• Completing the square for both the $x$ and $y$ terms is a crucial step in rewriting the equation in standard form.
• The center of the circle is given by the values of $h$ and $k$ in the standard form of the equation.
• The radius of the circle is given by the value of $r$ in the standard form of the equation.

Frequently Asked Questions

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

A: The equation of a circle in standard form 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.

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

A: To complete the square for the $x$ terms, we need to add $(8/2)^2 = 16$ to both sides of the equation. To complete the square for the $y$ terms, we need to add $(-6/2)^2 = 9$ to both sides of the equation.

Q: What is the center of the circle in the given equation?

A: The center of the circle in the given equation is $(-4, 3)$.

Q: What is the radius of the circle in the given equation?

A: The radius of the circle in the given equation is $\boxed{5}$ units.

Q: How do we find the radius of a circle from its equation?

A: To find the radius of a circle from its equation, we need to rewrite the equation in standard form and identify the value of $r$.

Q: What is the significance of completing the square in rewriting the equation of a circle?

A: Completing the square is a crucial step in rewriting the equation of a circle in standard form, as it allows us to identify the center and radius of the circle.

References

• [1] "Equation of a Circle" by Math Open Reference
• [2] "Completing the Square" by Khan Academy
• [3] "Standard Form of the Equation of a Circle" by Purplemath