Problem 46 - Algebra 2

Question

REASONING Explain why the graph of the equation \(x^{2}+y^{2}-4 x+2 y+5=0\) is a single point.

Step-by-Step Solution

Answer

The graph is a single point at \((2, -1)\) because the equation describes a circle with radius 0.

Step 1: Rearrange the Equation

Start with the given equation: \(x^2 + y^2 - 4x + 2y + 5 = 0\). Rearrange it to make it easier to complete the square. The equation becomes \(x^2 - 4x + y^2 + 2y = -5\).

Step 2: Complete the Square for x

Take the terms involving \(x\): \(x^2 - 4x\). To complete the square, take half of \(-4\), which is \(-2\), and square it to get \(4\). Add and subtract \(4\) inside the equation: \((x^2 - 4x + 4)\).

Step 3: Complete the Square for y

Now take the terms involving \(y\): \(y^2 + 2y\). To complete the square, take half of \(2\), which is \(1\), and square it to get \(1\). Add and subtract \(1\) inside the equation: \((y^2 + 2y + 1)\).

Step 4: Simplify the Equation

Now the equation is \((x-2)^2 - 4 + (y+1)^2 - 1 = -5\). Simplify it to get \((x-2)^2 + (y+1)^2 = 0\).

Step 5: Interpret the Completed Square Form

The equation \((x-2)^2 + (y+1)^2 = 0\) describes a circle with radius 0 and center at \((2, -1)\). A circle with radius 0 is actually a single point located exactly at the coordinates of the center.

Key Concepts

Quadratic Equation

Quadratic equations are polynomials of degree two, usually in the form of \(ax^2 + bx + c = 0\). These equations can describe parabolic shapes when graphed on a coordinate plane. However, they can also appear in more complex equations like conics.
In our original exercise, the equation starts as a quadratic involving both \(x\) and \(y\). It can be transformed to highlight specific points or lines when expressed by methods like completing the square. Completing the square is a technique that re-arranges a quadratic equation into a perfect square trinomial, making it easier to interpret its graph.
For instance, in the exercise solution, both the \(x\) and \(y\) terms were used to form squares: \((x-2)^2\) and \((y+1)^2\). This completed square form reveals much about the location and nature of the graph, turning a complex quadratic into a simple, recognizable shape.

Circle Equations

Circle equations are generally written as \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius of the circle.
In the exercise, by completing the square, the equation \((x-2)^2 + (y+1)^2 = 0\) was formed. Here, after completing the square, it becomes clear that the center is at \((2, -1)\), and the radius would be \(\sqrt{0} = 0\).
This radius of zero means the circle doesn't extend anywhere; it's just a point. This technique is a powerful way to transform and simplify equations to reveal the circle's center and radius directly, especially useful in more complex coordinate geometry problems.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves using algebraic methods to solve geometric problems. It provides a connection between geometric figures and algebraic equations.
In this exercise, coordinate geometry plays a crucial role. By manipulating the quadratic equation through techniques like completing the square, we translated a seemingly complex algebraic expression into a geometric concept - in this case, recognizing it as a circle with a radius of zero.
This blend of algebra and geometry allows for a deeper understanding of spatial relations and properties of shapes and is foundational in higher level mathematics. Through coordinate geometry, solving for intersections, distances, and locations becomes tangible and manageable.