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.