Consider A Circle Whose Equation Is X 2 + Y 2 + 4 X − 6 Y − 36 = 0 X^2+y^2+4x-6y-36=0 X 2 + Y 2 + 4 X − 6 Y − 36 = 0 . Which Statements Are True? Check All That Apply.- To Begin Converting The Equation To Standard Form, Subtract 36 From Both Sides.- To Complete The Square For The X X X Terms, Add

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. In this article, we will consider a circle whose equation is $x^2+y^2+4x-6y-36=0$ and determine which statements are true.

Converting the Equation to Standard Form

To convert the equation to standard form, we need to complete the square for both the $x$ and $y$ terms. However, before we do that, let's first subtract 36 from both sides of the equation.

Subtracting 36 from Both Sides

To begin converting the equation to standard form, subtract 36 from both sides. This statement is true. By subtracting 36 from both sides, we get $x^2+y^2+4x-6y=36$.

Completing the Square for the $x$ Terms

To complete the square for the $x$ terms, add $\left(\dfrac{4}{2}\right)^2=4$ to both sides. This statement is true. By adding 4 to both sides, we get $x^2+4x+4+y^2-6y=40$.

Completing the Square for the $y$ Terms

To complete the square for the $y$ terms, add $\left(\dfrac{-6}{2}\right)^2=9$ to both sides. This statement is true. By adding 9 to both sides, we get $x^2+4x+4+y^2-6y+9=49$.

Writing the Equation in Standard Form

Now that we have completed the square for both the $x$ and $y$ terms, we can write the equation in standard form.

$(x+2)^2+(y-3)^2=49$

This is the equation of a circle with center $(-2,3)$ and radius $\sqrt{49}=7$.

Conclusion

In this article, we considered a circle whose equation is $x^2+y^2+4x-6y-36=0$ and determined which statements are true. We found that the following statements are true:

• To begin converting the equation to standard form, subtract 36 from both sides.
• To complete the square for the $x$ terms, add $\left(\dfrac{4}{2}\right)^2=4$ to both sides.
• To complete the square for the $y$ terms, add $\left(\dfrac{-6}{2}\right)^2=9$ to both sides.

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

A: The standard form of 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.

Q: How do I convert the equation of a circle to standard form?

A: To convert the equation of a circle to standard form, you need to complete the square for both the $x$ and $y$ terms. This involves adding and subtracting the same value to both sides of the equation to create a perfect square trinomial.

Q: What is the first step in converting the equation of a circle to standard form?

A: The first step in converting the equation of a circle to standard form is to subtract the constant term from both sides of the equation. This will help to isolate the terms involving $x$ and $y$.

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

A: To complete the square for the $x$ terms, you need to add the square of half the coefficient of $x$ to both sides of the equation. This will create a perfect square trinomial.

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

A: To complete the square for the $y$ terms, you need to add the square of half the coefficient of $y$ to both sides of the equation. This will create a perfect square trinomial.

Q: What is the final step in converting the equation of a circle to standard form?

A: The final step in converting the equation of a circle to standard form is to write the equation in the form $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center of the circle and $r$ is the radius.

Q: How do I find the center and radius of a circle from its equation in standard form?

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

Q: What are some common mistakes to avoid when converting the equation of a circle to standard form?

A: Some common mistakes to avoid when converting the equation of a circle to standard form include:

• Not subtracting the constant term from both sides of the equation
• Not adding the square of half the coefficient of $x$ and $y$ to both sides of the equation
• Not writing the equation in the form $(x-h)^2 + (y-k)^2 = r^2$

Q: How can I practice converting the equation of a circle to standard form?

A: You can practice converting the equation of a circle to standard form by working through examples and exercises. You can also use online resources and practice problems to help you build your skills.

Q: What are some real-world applications of converting the equation of a circle to standard form?

A: Converting the equation of a circle to standard form has many real-world applications, including:

• Finding the center and radius of a circle in a coordinate plane
• Determining the distance between two points on a circle
• Calculating the area and circumference of a circle

By understanding how to convert the equation of a circle to standard form, you can apply this knowledge to a wide range of real-world problems and applications.