Change the equation in standard form of circle 1. x°+y+2V3x-б+3=0 2. x?+y-10x+6y+21=0 3. 9x2+92-12x-24y+20-9V5=0 4. 16x7+16y?+8x-48y-75=0

It seems like there are some typographical errors in the equations you've provided. I'll correct them to make sense of them and then convert them to the standard form of a circle's equation, which is:

\[
(x - h)^2 + (y - k)^2 = r^2
\]

where \((h, k)\) is the center of the circle, and \(r\) is the radius.

1. **Given equation:** \( x° + y + 2\sqrt{3}x - б + 3 = 0 \)

   Assuming the correct equation is \( x^2 + y + 2\sqrt{3}x - 6 + 3 = 0 \):
   
   **Corrected equation:** 
   \[
   x^2 + 2\sqrt{3}x + y - 3 = 0
   \]

   **Steps to convert to standard form:**
   - Group the \(x\) terms and complete the square: \( x^2 + 2\sqrt{3}x \) 
   - Completing the square: 
     \[
     x^2 + 2\sqrt{3}x + 3 = (x + \sqrt{3})^2
     \]
   - Equation becomes:
     \[
     (x + \sqrt{3})^2 + y = 3
     \]
   - Rewrite in standard form:
     \[
     (x + \sqrt{3})^2 + (y - 0)^2 = 3
     \]

   **Standard form:** 
   \[
   (x + \sqrt{3})^2 + y^2 = 3
   \]

2. **Given equation:** \( x? + y - 10x + 6y + 21 = 0 \)

   Assuming the correct equation is \( x^2 + y^2 - 10x + 6y + 21 = 0 \):

   **Steps to convert to standard form:**
   - Group the \(x\) terms and \(y\) terms: 
     \[
     x^2 - 10x + y^2 + 6y = -21
     \]
   - Complete the square for \(x\) and \(y\) terms:
     \[
     (x^2 - 10x + 25) + (y^2 + 6y + 9) = -21 + 25 + 9
     \]
     \[
     (x - 5)^2 + (y + 3)^2 = 13
     \]

   **Standard form:** 
   \[
   (x - 5)^2 + (y + 3)^2 = 13
   \]

3. **Given equation:** \( 9x^2 + 9y^2 - 12x - 24y + 20 - 9\sqrt{5} = 0 \)

   **Steps to convert to standard form:**
   - Divide the whole equation by 9:
     \[
     x^2 + y^2 - \frac{4}{3}x - \frac{8}{3}y + \frac{20}{9} - \sqrt{5} = 0
     \]
   - Move constants to the right side:
     \[
     x^2 - \frac{4}{3}x + y^2 - \frac{8}{3}y = \sqrt{5} - \frac{20}{9}
     \]
   - Complete the square:
     \[
     \left( x^2 - \frac{4}{3}x + \frac{4}{9} \right) + \left( y^2 - \frac{8}{3}y + \frac{16}{9} \right) = \sqrt{5} - \frac{20}{9} + \frac{20}{9}
     \]
     \[
     \left( x - \frac{2}{3} \right)^2 + \left( y - \frac{4}{3} \right)^2 = \sqrt{5}
     \]

   **Standard form:** 
   \[
   \left( x - \frac{2}{3} \right)^2 + \left( y - \frac{4}{3} \right)^2 = \sqrt{5}
   \]

4. **Given equation:** \( 16x^7 + 16y^2 + 8x - 48y - 75 = 0 \)

   Assuming the correct equation is \( 16x^2 + 16y^2 + 8x - 48y - 75 = 0 \):

   **Steps to convert to standard form:**
   - Divide the whole equation by 16:
     \[
     x^2 + y^2 + \frac{1}{2}x - 3y = \frac{75}{16}
     \]
   - Complete the square:
     \[
     \left( x^2 + \frac{1}{2}x + \frac{1}{16} \right) + \left( y^2 - 3y + \frac{9}{4} \right) = \frac{75}{16} + \frac{1}{16} + \frac{9}{4}
     \]
     \[
     \left( x + \frac{1}{4} \right)^2 + \left( y - \frac{3}{2} \right)^2 = \frac{100}{16}
     \]

   **Standard form:** 
   \[
   \left( x + \frac{1}{4} \right)^2 + \left( y - \frac{3}{2} \right)^2 = \frac{25}{4}
   \]

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Would you like more details or have any questions about the process? Here are some related questions you might consider:

1. How do you complete the square for any quadratic equation?
2. What are the characteristics of a circle's equation in standard form?
3. How do you determine the center and radius from the standard form equation?
4. Can you explain how to simplify a circle's equation by dividing it by a constant?
5. What is the geometric significance of completing the square in circle equations?

**Tip:** Always check for typographical errors in mathematical equations before solving them to ensure accuracy.

Learn how to convert various algebraic equations into the standard form of a circle's equation. This detailed guide covers the process of completing the square, identifying the center and radius, and correcting typographical errors. Suitable for students in Grades 9-11.

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