Which Equations Represent Circles That Have A Diameter Of 12 Units And A Center That Lies On The $y$-axis? Choose Two Correct Answers.A. $x 2+(y-5) 2=6$B. $x 2+(y+8) 2=36$C. $x 2+(y-3) 2=36$D.

by ADMIN 201 views

Circles are a fundamental concept in mathematics, and their equations play a crucial role in geometry and trigonometry. In this article, we will explore the equations that represent circles with a diameter of 12 units and a center that lies on the $y$-axis.

What is a Circle?

A circle is a set of points that are equidistant from a central point, known as the center. The distance from the center to any point on the circle is called the radius. The diameter of a circle is twice the radius and represents the longest distance across the circle.

Equations of Circles

The general equation of a circle with center $(h, k)$ and radius $r$ is given by:

(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

This equation represents a circle with center $(h, k)$ and radius $r$.

Circles with a Diameter of 12 Units

Since the diameter of the circle is 12 units, the radius is half of the diameter, which is 6 units. We need to find the equations of circles with a radius of 6 units and a center that lies on the $y$-axis.

Equation A: $x2+(y-5)2=6$

Let's analyze Equation A:

x2+(y−5)2=6x^2+(y-5)^2=6

In this equation, the center of the circle is $(0, 5)$, which lies on the $y$-axis. The radius of the circle is $\sqrt{6}$ units, which is less than 6 units. Therefore, Equation A does not represent a circle with a diameter of 12 units.

Equation B: $x2+(y+8)2=36$

Now, let's analyze Equation B:

x2+(y+8)2=36x^2+(y+8)^2=36

In this equation, the center of the circle is $(0, -8)$, which lies on the $y$-axis. The radius of the circle is $\sqrt{36} = 6$ units, which matches the given diameter. Therefore, Equation B represents a circle with a diameter of 12 units and a center that lies on the $y$-axis.

Equation C: $x2+(y-3)2=36$

Finally, let's analyze Equation C:

x2+(y−3)2=36x^2+(y-3)^2=36

In this equation, the center of the circle is $(0, 3)$, which lies on the $y$-axis. The radius of the circle is $\sqrt{36} = 6$ units, which matches the given diameter. Therefore, Equation C represents a circle with a diameter of 12 units and a center that lies on the $y$-axis.

Conclusion

In conclusion, the two correct answers are Equation B and Equation C. Both equations represent circles with a diameter of 12 units and a center that lies on the $y$-axis.

Key Takeaways

  • The general equation of a circle with center $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$.
  • The radius of a circle is half of the diameter.
  • The center of a circle lies on the $y$-axis if the $x$-coordinate is 0.

Frequently Asked Questions

Q: What is the equation of a circle with a diameter of 12 units and a center that lies on the $y$-axis?

A: The equation of a circle with a diameter of 12 units and a center that lies on the $y$-axis is given by $(x - h)^2 + (y - k)^2 = 6^2$, where $(h, k)$ is the center of the circle.

Q: How do I find the center of a circle from its equation?

A: To find the center of a circle from its equation, you need to identify the values of $h$ and $k$ in the equation $(x - h)^2 + (y - k)^2 = r^2$. The center of the circle is given by $(h, k)$.

Q: What is the radius of a circle with a diameter of 12 units?

A: The radius of a circle with a diameter of 12 units is half of the diameter, which is 6 units.

References

In our previous article, we explored the equations that represent circles with a diameter of 12 units and a center that lies on the $y$-axis. In this article, we will delve deeper into the world of circle equations and provide a comprehensive Q&A guide.

Q&A: Circle Equations

Q: What is the general equation of a circle?

A: The general equation of a circle with center $(h, k)$ and radius $r$ is given by:

(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

Q: How do I find the center of a circle from its equation?

A: To find the center of a circle from its equation, you need to identify the values of $h$ and $k$ in the equation $(x - h)^2 + (y - k)^2 = r^2$. The center of the circle is given by $(h, k)$.

Q: What is the radius of a circle with a diameter of 12 units?

A: The radius of a circle with a diameter of 12 units is half of the diameter, which is 6 units.

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

A: To find the equation of a circle with a given center and radius, you can use the general equation of a circle:

(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

Replace $(h, k)$ with the given center and $r$ with the given radius.

Q: What is the equation of a circle with a center at $(0, 5)$ and a radius of 6 units?

A: The equation of a circle with a center at $(0, 5)$ and a radius of 6 units is given by:

(x−0)2+(y−5)2=62(x - 0)^2 + (y - 5)^2 = 6^2

Simplifying the equation, we get:

x2+(y−5)2=36x^2 + (y - 5)^2 = 36

Q: What is the equation of a circle with a center at $(0, -8)$ and a radius of 6 units?

A: The equation of a circle with a center at $(0, -8)$ and a radius of 6 units is given by:

(x−0)2+(y+8)2=62(x - 0)^2 + (y + 8)^2 = 6^2

Simplifying the equation, we get:

x2+(y+8)2=36x^2 + (y + 8)^2 = 36

Q: What is the equation of a circle with a center at $(0, 3)$ and a radius of 6 units?

A: The equation of a circle with a center at $(0, 3)$ and a radius of 6 units is given by:

(x−0)2+(y−3)2=62(x - 0)^2 + (y - 3)^2 = 6^2

Simplifying the equation, we get:

x2+(y−3)2=36x^2 + (y - 3)^2 = 36

Common Mistakes to Avoid

  • Incorrectly identifying the center of a circle: Make sure to identify the values of $h$ and $k$ in the equation $(x - h)^2 + (y - k)^2 = r^2$ to find the center of the circle.
  • Incorrectly calculating the radius: Make sure to calculate the radius correctly by dividing the diameter by 2.
  • Not simplifying the equation: Make sure to simplify the equation to get the correct form.

Tips and Tricks

  • Use the general equation of a circle: The general equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius.
  • Identify the center and radius: Identify the values of $h$ and $k$ in the equation to find the center of the circle, and identify the value of $r$ to find the radius.
  • Simplify the equation: Simplify the equation to get the correct form.

Conclusion

In conclusion, the general equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. By following the tips and tricks provided in this article, you can easily find the equation of a circle with a given center and radius.

Frequently Asked Questions

Q: What is the equation of a circle with a center at $(0, 5)$ and a radius of 6 units?

A: The equation of a circle with a center at $(0, 5)$ and a radius of 6 units is given by:

(x−0)2+(y−5)2=62(x - 0)^2 + (y - 5)^2 = 6^2

Simplifying the equation, we get:

x2+(y−5)2=36x^2 + (y - 5)^2 = 36

Q: What is the equation of a circle with a center at $(0, -8)$ and a radius of 6 units?

A: The equation of a circle with a center at $(0, -8)$ and a radius of 6 units is given by:

(x−0)2+(y+8)2=62(x - 0)^2 + (y + 8)^2 = 6^2

Simplifying the equation, we get:

x2+(y+8)2=36x^2 + (y + 8)^2 = 36

Q: What is the equation of a circle with a center at $(0, 3)$ and a radius of 6 units?

A: The equation of a circle with a center at $(0, 3)$ and a radius of 6 units is given by:

(x−0)2+(y−3)2=62(x - 0)^2 + (y - 3)^2 = 6^2

Simplifying the equation, we get:

x2+(y−3)2=36x^2 + (y - 3)^2 = 36

References