To analyze and solve the equation $x^2 + y^2 - 8x + 6y - 11 = 0$, we will rewrite it in standard form by completing the square.

Step 1: Group the terms

Group $x$-terms and $y$-terms together: $(x^2 - 8x) + (y^2 + 6y) = 11$

Step 2: Complete the square

For $x^2 - 8x$:

Take half the coefficient of $x$ (which is $-8$), square it, and add and subtract the square inside the parentheses: $x^2 - 8x = (x - 4)^2 - 16$

For $y^2 + 6y$:

Take half the coefficient of $y$ (which is $6$), square it, and add and subtract the square: $y^2 + 6y = (y + 3)^2 - 9$

Step 3: Substitute back into the equation

$(x - 4)^2 - 16 + (y + 3)^2 - 9 = 11$

Combine constants: $(x - 4)^2 + (y + 3)^2 - 25 = 11$

$(x - 4)^2 + (y + 3)^2 = 36$

Step 4: Interpret the equation

This is the equation of a circle in standard form: $(x - h)^2 + (y - k)^2 = r^2$ where $(h, k)$ is the center and $r$ is the radius.

• Center: $(4, -3)$
• Radius: $r = \sqrt{36} = 6$

Final Answer:

The given equation represents a circle with:

• Center: $(4, -3)$
• Radius: $6$

Would you like further clarification or details on any step?

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Related Questions:

1. How do you complete the square for other types of equations?
2. Can you show an example where the radius is zero, making it a point?
3. What is the general form of a circle equation?
4. How do you graph a circle based on its standard form?
5. How do translations of the center affect the equation of a circle?

Tip: Always double-check your arithmetic when completing the square to avoid errors in the standard form equation.