Type The Correct Answer In Each Box.The General Form Of The Equation Of A Circle Is $x^2+y^2+8x+22y+37=0$.The Equation Of This Circle In Standard Form Is \$(x+$ $\square$ $)^2+(y+$ $\square$

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Understanding the General Form of a Circle Equation

The general form of a circle equation is given by $x2+y2+Ax+By+C=0$, where AA, BB, and CC are constants. In this case, the equation of the circle is $x2+y2+8x+22y+37=0$. To convert this equation to standard form, we need to complete the square for both the xx and yy terms.

Completing the Square for the x Terms

To complete the square for the xx terms, we need to add and subtract (A2)2(\frac{A}{2})^2 inside the parentheses. In this case, A=8A=8, so we need to add and subtract (82)2=16(\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(x+4)^2. This is the completed square for the xx terms.

Completing the Square for the y Terms

To complete the square for the yy terms, we need to add and subtract (B2)2(\frac{B}{2})^2 inside the parentheses. In this case, B=22B=22, so we need to add and subtract (222)2=121(\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(y+11)^2. This is the completed square for the yy terms.

Converting the Equation to Standard Form

Now that we have completed the square for both the xx and yy 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(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 xx and yy 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(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 $x2+y2+6x+12y+36=0$. Convert this equation to standard form.

Solution

To convert this equation to standard form, we need to complete the square for both the xx and yy terms. We have A=6A=6 and B=12B=12, so we need to add and subtract (62)2=9(\frac{6}{2})^2=9 and (122)2=36(\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(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 $x2+y2-4x-6y-5=0$. Convert this equation to standard form.

Solution

To convert this equation to standard form, we need to complete the square for both the xx and yy terms. We have A=−4A=-4 and B=−6B=-6, so we need to add and subtract (−42)2=4(\frac{-4}{2})^2=4 and (−62)2=9(\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(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 $x2+y2+2x+4y+4=0$. Convert this equation to standard form.

Solution

To convert this equation to standard form, we need to complete the square for both the xx and yy terms. We have A=2A=2 and B=4B=4, so we need to add and subtract (22)2=1(\frac{2}{2})^2=1 and (42)2=4(\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(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 $x2+y2+10x+20y+100=0$. Convert this equation to standard form.

Problem 2

The equation of a circle is given by $x2+y2-8x-16y-64=0$. Convert this equation to standard form.

Problem 3

The equation of a circle is given by $x2+y2+6x+12y+36=0$. 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 $x2+y2+Ax+By+C=0$, where AA, BB, and CC 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 xx and yy terms. This involves adding and subtracting (A2)2(\frac{A}{2})^2 and (B2)2(\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(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 xx terms, you need to add and subtract (A2)2(\frac{A}{2})^2 inside the parentheses. For example, if A=8A=8, you would add and subtract (82)2=16(\frac{8}{2})^2=16.

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

A: To complete the square for the yy terms, you need to add and subtract (B2)2(\frac{B}{2})^2 inside the parentheses. For example, if B=22B=22, you would add and subtract (222)2=121(\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)(h,k) is the center of the circle and rr 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:

  1. Complete the square for the xx terms by adding and subtracting (A2)2(\frac{A}{2})^2 inside the parentheses.
  2. Complete the square for the yy terms by adding and subtracting (B2)2(\frac{B}{2})^2 inside the parentheses.
  3. Add and subtract the constants inside the parentheses.
  4. 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 $x2+y2+8x+22y+37=0$. Convert this equation to standard form.
  • The equation of a circle is given by $x2+y2-4x-6y-5=0$. Convert this equation to standard form.
  • The equation of a circle is given by $x2+y2+6x+12y+36=0$. 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 $x2+y2+10x+20y+100=0$. Convert this equation to standard form.
  • The equation of a circle is given by $x2+y2-8x-16y-64=0$. Convert this equation to standard form.
  • The equation of a circle is given by $x2+y2+6x+12y+36=0$. 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.