Problem 54 - Precalculus

Question

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$x^{2}+y^{2}-4 x-6 y-23=0$$

Step-by-Step Solution

Answer

The given equation, after performing completing the square and simplification, represents a circle with center at (2, 3) and radius \( \sqrt{7}\).

Step 1: Rearrange and group the equation

Rearrange the equation as \(x^{2}-4x+y^{2}-6y=-23\). Group the x and y terms separately: \((x^2-4x)+(y^2-6y)=-23\).

Step 2: Complete the squares

To complete the squares, add the square of half the coefficient of x and y on both sides of the equation: \((x^2-4x+4)+(y^2-6y+9)=-23+4+9\). The equation now can be written as \((x-2)^2+(y-3)^2=30-23\).

Step 3: Identify the conic section

The equation is now in the form \((x-h)^2+(y-k)^2=r^2\), which is the equation of a circle where (h,k) is the center and r is the radius. In this case, the center of the circle is at (2, 3) and the radius is \( \sqrt{7}\).

Key Concepts

Completing the Square

One of the most important algebraic techniques used in solving equations and analyzing graphs of conic sections is completing the square. This method allows us to rewrite quadratic expressions as perfect squares, making them easier to interpret and solve.

For example, consider the quadratic portion of the conic section equation from our exercise: the terms with the variables x and y. To complete the square for the term x² – 4x, we take half the coefficient of x, which is -4, divide by 2 to get -2, and then square it to receive 4. This value is then added and subtracted inside the parentheses to maintain balance in the equation. The same process is applied to the term with y, resulting in the addition of 9 to both sides of the equation.

By completing the square, we convert the equation into a form where it represents a perfect square trinomial (x – h)² and (y – k)², which are easier to analyze and are pivotal in identifying the type of conic section they represent.

Circle Equation

Analyzing circles in coordinate geometry often involves the use of the standard circle equation, which has the form (x – h)² + (y – k)² = r². Here, (h, k) represents the center of the circle, and r is the radius.

In our exercise, after completing the square, we arrived at the equation (x – 2)² + (y – 3)² = 7. This clearly fits the standard form of a circle's equation, indicating that the graph is indeed a circle. We can interpret (2, 3) as the center of the circle and √7 as the radius. Understanding this formula allows students to quickly visualize and graph circles, as well as to solve problems related to the properties of circles.

Conic Section Classification

Conic sections are the curves formed by the intersection of a plane with a double napped cone. They include circles, ellipses, parabolas, and hyperbolas. The general second-degree equation Ax² + Bxy + Cy² + Dx + Ey + F = 0 can represent any of these conic sections depending on the coefficients and their relationships.

• For a circle, A = C and B = 0.
• An ellipse has A ≠ C, with B = 0 and AC > 0.
• A parabola will have either A or C equal to 0, but not both.
• Lastly, a hyperbola is defined by A ≠ C, with B = 0 and AC < 0.

By completing the square and reorganizing terms as we did in the step-by-step solution, we simplify the equation to a form that makes classification straightforward. For the given exercise, since the equation resulted in a format that corresponds solely to a circle, we successfully classified the graph without ambiguity.