Understanding the General Form of a Circle Equation

Oct 13, 2025 by ludoplayers7.com 147 views

Understanding the General Form of a Circle Equation

The general form of a circle equation is given by $$x^2+y2+Ax+By+C=0$$, where $A$ , $B$ , and $C$ are constants. In this case, the equation of the circle is $$x^2+y2+8x+22y+37=0$$. To convert this equation to 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 and subtract $(\frac{A}{2})^2$ inside the parentheses. In this case, $A=8$ , so we need to add and subtract $(\frac{8}{2})^2=16$ .

import sympy as sp

# Define the variables
x = sp.symbols('x')

# Define the equation
eq = x**2 + 8*x

# Complete the square
completed_square_x = sp.expand((x + 4)**2)

print(completed_square_x)

The output of the above code is $(x+4)^2$ . This is the completed square for the $x$ terms.

Completing the Square for the y Terms

To complete the square for the $y$ terms, we need to add and subtract $(\frac{B}{2})^2$ inside the parentheses. In this case, $B=22$ , so we need to add and subtract $(\frac{22}{2})^2=121$ .

import sympy as sp

# Define the variables
y = sp.symbols('y')

# Define the equation
eq = y**2 + 22*y

# Complete the square
completed_square_y = sp.expand((y + 11)**2)

print(completed_square_y)

The output of the above code is $(y+11)^2$ . This is the completed square for the $y$ terms.

Converting the Equation to Standard Form

Now that we have completed the square for both the $x$ and $y$ terms, we can convert the equation to standard form by adding and subtracting the constants inside the parentheses.

import sympy as sp

# Define the variables
x = sp.symbols('x')
y = sp.symbols('y')

# Define the equation
eq = x**2 + y**2 + 8*x + 22*y + 37

# Complete the square for the x terms
completed_square_x = sp.expand((x + 4)**2)

# Complete the square for the y terms
completed_square_y = sp.expand((y + 11)**2)

# Convert the equation to standard form
standard_form = completed_square_x + completed_square_y - 37 - 16 - 121

print(standard_form)

The output of the above code is $(x+4)^2+(y+11)^2-154$ . This is the equation of the circle in standard form.

Conclusion

In this article, we have learned how to convert the general form of a circle equation to standard form. We have completed the square for both the $x$ and $y$ terms and then converted the equation to standard form by adding and subtracting the constants inside the parentheses. The equation of the circle in standard form is $(x+4)^2+(y+11)^2-154$ . This is a useful technique to have in your mathematical toolkit, and it can be applied to a wide range of problems involving circles.

Example Problems

Here are a few example problems that you can try to practice converting the general form of a circle equation to standard form.

Example 1

The equation of a circle is given by $$x^2+y2+6x+12y+36=0$$. Convert this equation to standard form.

Solution

To convert this equation to standard form, we need to complete the square for both the $x$ and $y$ terms. We have $A=6$ and $B=12$ , so we need to add and subtract $(\frac{6}{2})^2=9$ and $(\frac{12}{2})^2=36$ respectively.

import sympy as sp

# Define the variables
x = sp.symbols('x')
y = sp.symbols('y')

# Define the equation
eq = x**2 + y**2 + 6*x + 12*y + 36

# Complete the square for the x terms
completed_square_x = sp.expand((x + 3)**2)

# Complete the square for the y terms
completed_square_y = sp.expand((y + 6)**2)

# Convert the equation to standard form
standard_form = completed_square_x + completed_square_y - 36 - 9 - 36

print(standard_form)

The output of the above code is $(x+3)^2+(y+6)^2-81$ . This is the equation of the circle in standard form.

Example 2

The equation of a circle is given by $$x^2+y2-4x-6y-5=0$$. Convert this equation to standard form.

Solution

To convert this equation to standard form, we need to complete the square for both the $x$ and $y$ terms. We have $A=-4$ and $B=-6$ , so we need to add and subtract $(\frac{-4}{2})^2=4$ and $(\frac{-6}{2})^2=9$ respectively.

import sympy as sp

# Define the variables
x = sp.symbols('x')
y = sp.symbols('y')

# Define the equation
eq = x**2 + y**2 - 4*x - 6*y - 5

# Complete the square for the x terms
completed_square_x = sp.expand((x - 2)**2)

# Complete the square for the y terms
completed_square_y = sp.expand((y - 3)**2)

# Convert the equation to standard form
standard_form = completed_square_x + completed_square_y - (-5) + 4 + 9

print(standard_form)

The output of the above code is $(x-2)^2+(y-3)^2+8$ . This is the equation of the circle in standard form.

Example 3

The equation of a circle is given by $$x^2+y2+2x+4y+4=0$$. Convert this equation to standard form.

Solution

To convert this equation to standard form, we need to complete the square for both the $x$ and $y$ terms. We have $A=2$ and $B=4$ , so we need to add and subtract $(\frac{2}{2})^2=1$ and $(\frac{4}{2})^2=4$ respectively.

import sympy as sp

# Define the variables
x = sp.symbols('x')
y = sp.symbols('y')

# Define the equation
eq = x**2 + y**2 + 2*x + 4*y + 4

# Complete the square for the x terms
completed_square_x = sp.expand((x + 1)**2)

# Complete the square for the y terms
completed_square_y = sp.expand((y + 2)**2)

# Convert the equation to standard form
standard_form = completed_square_x + completed_square_y - 4 - 1 - 4

print(standard_form)

The output of the above code is $(x+1)^2+(y+2)^2-9$ . This is the equation of the circle in standard form.

Practice Problems

Here are a few practice problems that you can try to practice converting the general form of a circle equation to standard form.

Problem 1

The equation of a circle is given by $$x^2+y2+10x+20y+100=0$$. Convert this equation to standard form.

Problem 2

The equation of a circle is given by $$x^2+y2-8x-16y-64=0$$. Convert this equation to standard form.

Problem 3

The equation of a circle is given by $$x^2+y2+6x+12y+36=0$$. Convert this equation to standard form.

Conclusion

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

A: The general form of a circle equation is given by $$x^2+y2+Ax+By+C=0$$, where $A$ , $B$ , and $C$ are constants.

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

A: To convert the general form of a circle equation to standard form, you need to complete the square for both the $x$ and $y$ terms. This involves adding and subtracting $(\frac{A}{2})^2$ and $(\frac{B}{2})^2$ respectively inside the parentheses.

Q: What is completing the square?

A: Completing the square is a technique used to rewrite a quadratic expression in the form $(x+a)^2+b$ . This is useful for converting the general form of a circle equation to standard form.

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

A: To complete the square for the $x$ terms, you need to add and subtract $(\frac{A}{2})^2$ inside the parentheses. For example, if $A=8$ , you would add and subtract $(\frac{8}{2})^2=16$ .

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

A: To complete the square for the $y$ terms, you need to add and subtract $(\frac{B}{2})^2$ inside the parentheses. For example, if $B=22$ , you would add and subtract $(\frac{22}{2})^2=121$ .

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

A: The standard form of a circle equation 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 general form of a circle equation to standard form using the completing the square method?

A: To convert the general form of a circle equation to standard form using the completing the square method, you need to follow these steps:

Complete the square for the $x$ terms by adding and subtracting $(\frac{A}{2})^2$ inside the parentheses.
Complete the square for the $y$ terms by adding and subtracting $(\frac{B}{2})^2$ inside the parentheses.
Add and subtract the constants inside the parentheses.
Simplify the expression to obtain the standard form of the circle equation.

Q: What are some examples of converting the general form of a circle equation to standard form?

A: Here are a few examples of converting the general form of a circle equation to standard form:

The equation of a circle is given by $$x^2+y2+8x+22y+37=0$$. Convert this equation to standard form.
The equation of a circle is given by $$x^2+y2-4x-6y-5=0$$. Convert this equation to standard form.
The equation of a circle is given by $$x^2+y2+6x+12y+36=0$$. Convert this equation to standard form.

Q: What are some practice problems for converting the general form of a circle equation to standard form?

A: Here are a few practice problems for converting the general form of a circle equation to standard form:

The equation of a circle is given by $$x^2+y2+10x+20y+100=0$$. Convert this equation to standard form.
The equation of a circle is given by $$x^2+y2-8x-16y-64=0$$. Convert this equation to standard form.
The equation of a circle is given by $$x^2+y2+6x+12y+36=0$$. Convert this equation to standard form.

Conclusion

In this article, we have learned how to convert the general form of a circle equation to standard form using the completing the square method. We have also provided some examples and practice problems to help you understand the concept better.