To rewrite the given equation of the circle $x^2 + y^2 - 12x - 10y + 56 = 0$ in standard form using the method of completing the square, follow these steps:

Step 1: Rearrange the terms.

Group the $x$-terms and $y$-terms together and move the constant to the other side of the equation: $x^2 - 12x + y^2 - 10y = -56$

Step 2: Complete the square for the $x$-terms.

Take the coefficient of $x$, divide it by 2, and then square it: $\text{For } x^2 - 12x: \left(\frac{-12}{2}\right)^2 = (-6)^2 = 36$ Now add 36 to both sides of the equation: $x^2 - 12x + 36 + y^2 - 10y = -56 + 36$ $(x - 6)^2 + y^2 - 10y = -20$

Step 3: Complete the square for the $y$-terms.

Take the coefficient of $y$, divide it by 2, and then square it: $\text{For } y^2 - 10y: \left(\frac{-10}{2}\right)^2 = (-5)^2 = 25$ Now add 25 to both sides of the equation: $(x - 6)^2 + (y - 5)^2 = -20 + 25$ $(x - 6)^2 + (y - 5)^2 = 5$

Step 4: Write the equation in standard form.

The equation of the circle is now in standard form: $(x - 6)^2 + (y - 5)^2 = 5$ This represents a circle with center $(6, 5)$ and radius $\sqrt{5}$.

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Related questions:

1. How can you derive the general formula for completing the square?
2. How do you interpret the center and radius of a circle from its standard form?
3. Can you apply this method to a circle equation with different coefficients?
4. What are the geometric implications of completing the square?
5. How would this process change for an ellipse or hyperbola?

Tip:

When completing the square, always remember to balance the equation by adding the same value to both sides to maintain equality.