Determining the Value of k for a Circle Representation in Equations

When dealing with algebraic equations, determining the value of k for which the equation represents a circle is a fundamental concept in mathematics. This article explores how to find the appropriate value of k for the given equation 2x2 2y2 - 6x - 4y k 0. We will use step-by-step algebraic manipulation to transform the equation into the standard form of a circle equation.

Transformation of the Equation

To start, let's transform the given equation into a more manageable format. The initial equation is:

2x2 2y2 - 6x - 4y k 0

First, we divide the entire equation by 2 to simplify:

x2 y2 - 3x - 2y frac{k}{2} 0

Next, we aim to complete the square for both the x and y terms. Starting with the x terms:

x2 - 3x left(frac{3}{2}right)^2 - left(frac{3}{2}right)^2

And for the y terms:

y2 - 2y 1 - 1

Combining these, the equation becomes:

left(x - frac{3}{2}right)^2 (y - 1)^2 frac{9}{4} - frac{k}{2} 1

Further simplification yields:

left(x - frac{3}{2}right)^2 (y - 1)^2 frac{13 - 2k}{4}

This equation represents a circle if and only if the right-hand side is non-negative:

frac{13 - 2k}{4} geq 0

Thus, solving for k:

13 - 2k geq 0 2k leq 13 k leq frac{13}{2}

Center and Radius of the Circle

The center of the circle, given by the transformed equation, is:

left(-frac{3}{2}, 1right)

The radius of the circle, derived from the right-hand side, is:

r sqrt{frac{13 - 2k}{4}}

To ensure the radius is non-negative, we have:

frac{13 - 2k}{4} geq 0

This confirms that the provided value of k must satisfy:

k leq frac{13}{2}

Conclusion

In conclusion, the value of k for the given equation to represent a circle is any real number less than or equal to (frac{13}{2}). The center of the circle is located at left(-frac{3}{2}, 1right)) and the radius is given by (sqrt{frac{13 - 2k}{4}}).

Key Takeaways:

Determine the value of k for which a given algebraic equation represents a circle. Transform the given equation using algebraic manipulation and completing the square. Ensure the right-hand side of the circle equation is non-negative.

Understanding these steps is crucial for solving similar problems in algebra and geometry.