Problem 32 - Algebra and Trigonometry Real Mathematics, Real People

Question

Sketch the circle. Identify its center and radius. $$x^{2}+6 x+y^{2}-12 y+41=0$$

Step-by-Step Solution

Answer

The center of the circle is (-3, 6) and the radius is 2.

Step 1: Complete the square

To put our equation in standard form, we complete the square for the x and y terms. Our equation is \(x^{2}+6 x+y^{2}-12 y+41=0\). Split it into two separate squares. We get \(x^{2}+6 x + y^{2}-12 y = -41\). Find the number needed to complete the square for both x and y. For x, the number is \((6/2)^{2}=9\) and for y, the number is \((-12/2)^{2}=36\).

Step 2: Convert to standard form

Now convert our equation to standard form by adding the completing squares we calculated and moving the constant to the other side. This gives us \((x^{2}+6 x + 9) + (y^{2}-12 y + 36) = -41+9+36\). Simplify this to \((x+3)^{2} + (y-6)^{2} = 4\).

Step 3: Identify the center and radius

From this equation in standard form, we can identify the center of the circle as \((-3, 6)\) and the radius as \(\sqrt{4} = 2\).

Key Concepts

Completing the Square

Completing the square is a vital technique used to transform quadratic equations into a form that makes it easier to visually interpret important features of the function, such as the vertex of a parabola or, as in our case, the center and radius of a circle. It involves converting a quadratic expression into a perfect square trinomial, which is an expression that can be factored into a binomial squared.

For example, given a quadratic expression like \( ax^2 + bx + c \), the process would involve isolating the \( x^2 \) and \( x \) terms, then adding and subtracting the square of half the coefficient of \( x \), which is \( (b/2)^2 \). We apply this process separately to both the \( x \) and \( y \) terms in the equation of a circle, reorganizing it into the standard form and revealing the circle's geometric properties.

Standard Form of a Circle

The standard form of a circle's equation is one that provides a clear picture of the circle's properties, specifically its center and radius. This form is given as \( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h, k) \) is the center of the circle and \( r \) is its radius. The process of completing the square, as described earlier, is used to manipulate the general quadratic form of a circle into this standard form. This can be thought of as ‘unpacking’ the squared terms to reveal the location of the circle in the Cartesian coordinate system, as well as its size. It’s akin to finding the DNA of the circle — the foundational information from which all else about the circle’s geometry can be deduced.

Circle Center and Radius

Identifying the center and radius of a circle from its equation in standard form is a straightforward process. The center \( (h, k) \) is found directly from the transformed equation. These are the constants in the binomial squares which offset the \( x \) and \( y \) variables. As shown in the solution steps, the sign of these constants is reversed in the standard equation. The radius \( r \) is simply the square root of the constant term that completes the equation after the binomials have been squared. Learning to identify these components quickly can transform a daunting algebraic expression into a simple, geometric shape in your mind’s eye — bridging the gap between abstract algebra and concrete geometry.

Equation of a Circle

The equation of a circle in its general form may not immediately disclose much about the circle's characteristics, but through completing the square, it can be reshaped into the standard form that clearly defines the position and size of the circle. As we've seen, completing the square unwraps the equations to show the center at \( (h, k) \) and the square of the radius, \( r^2 \). Understanding how to derive the equation of a circle and interpreting its standard form is essential, as it provides a link between algebraic equations and geometric shapes, enabling us to sketch the circle accurately on a coordinate plane. It’s a perfect demonstration of the synergy between algebra and geometry — both necessary to master for a well-rounded mathematical education.