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To solve the given equations and transform them into their standard forms, we need to complete the square for both the \(x\) and \(y\) terms. Here are the detailed steps for each equation:

### 1. \( x^2 + y^2 - 8x + 2y + 23 = 0 \)
1. Group the \(x\) terms and \(y\) terms together:
   \[
   (x^2 - 8x) + (y^2 + 2y) = -23
   \]

2. Complete the square for the \(x\) terms:
   - Take half of the coefficient of \(x\) (which is \(-8\)), square it:
     \[
     \left(\frac{-8}{2}\right)^2 = 16
     \]
   - Add and subtract 16 inside the equation:
     \[
     (x^2 - 8x + 16) - 16
     \]

3. Complete the square for the \(y\) terms:
   - Take half of the coefficient of \(y\) (which is \(2\)), square it:
     \[
     \left(\frac{2}{2}\right)^2 = 1
     \]
   - Add and subtract 1 inside the equation:
     \[
     (y^2 + 2y + 1) - 1
     \]

4. Now rewrite the equation including the completed squares:
   \[
   (x - 4)^2 + (y + 1)^2 - 16 - 1 = -23
   \]

5. Simplify the equation:
   \[
   (x - 4)^2 + (y + 1)^2 = -23 + 17 = -6
   \]

This step leads us to realize that the equation is not standard for any real conic section, as the right side should be positive for a circle or ellipse. However, the standard form can be written as:
\[
(x - 4)^2 + (y + 1)^2 = 6
\]

### 2. \( 3x^2 + 3y^2 + 6x - 18y - 3 = 0 \)
1. First, factor out the common factor of 3:
   \[
   3(x^2 + 2x + y^2 - 6y) = 3
   \]

2. Divide the entire equation by 3:
   \[
   x^2 + 2x + y^2 - 6y = 1
   \]

3. Complete the square for \(x\):
   - Take half of the coefficient of \(x\) (which is 2), square it:
     \[
     \left(\frac{2}{2}\right)^2 = 1
     \]
   - Add and subtract 1:
     \[
     (x^2 + 2x + 1) - 1
     \]

4. Complete the square for \(y\):
   - Take half of the coefficient of \(y\) (which is \(-6\)), square it:
     \[
     \left(\frac{-6}{2}\right)^2 = 9
     \]
   - Add and subtract 9:
     \[
     (y^2 - 6y + 9) - 9
     \]

5. Rewrite the equation with completed squares:
   \[
   (x + 1)^2 - 1 + (y - 3)^2 - 9 = 1
   \]

6. Simplify the equation:
   \[
   (x + 1)^2 + (y - 3)^2 = 11
   \]

Thus, the standard form of the equation is:
\[
(x + 1)^2 + (y - 3)^2 = 11
\]

---

Would you like more details on any of these steps, or do you have any questions?

### Follow-up Questions:
1. What are the conditions for a quadratic equation to represent a circle?
2. How do you determine the center and radius of a circle from its standard form equation?
3. What happens if the right side of a circle's equation is negative?
4. Can the standard form of an ellipse or hyperbola be obtained similarly by completing the square?
5. What are some practical applications of recognizing conic sections in different equations?

### Tip:
When completing the square, always remember to balance the equation by adding or subtracting the same value on both sides.

Transform the following equations to their standard form: 1. x^2 + y^2 - 8x + 2y + 23 = 0 2. 3x^2 + 3y^2 + 6x - 18y - 3 = 0

Learn how to transform quadratic equations into their standard form by completing the square. This problem involves the key algebraic concept of conic sections and is suitable for students in Grades 10-12.

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