Problem 53 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 47-56: Find the center and radius of the circle with the given equation. $$ x^{2}+y^{2}+4 x-2 y+5=0 $$

Step-by-Step Solution

Answer

The center is (-2, 1) and the radius is 0.

Step 1: Rewrite the Circle Equation

The equation of a circle is typically written in the form \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. We need to complete the square for both \(x\) and \(y\) terms to rewrite the given equation into this form.

Step 2: Complete the Square for x

Start with the terms involving \(x\): \(x^2 + 4x\). To complete the square, take half of the coefficient of \(x\), which is 4, divide by 2 to get 2, then square it to get 4. Add and subtract 4 from the equation, leading to: \(x^2 + 4x + 4 - 4\). This can be rewritten as \((x+2)^2 - 4\).

Step 3: Complete the Square for y

Now, look at the \(y\) terms: \(y^2 - 2y\). Take half of the coefficient of \(y\), which is -2, divide by 2 to get -1, then square it to get 1. Add and subtract 1 from the equation, resulting in: \(y^2 - 2y + 1 - 1\). This can be rewritten as \((y-1)^2 - 1\).

Step 4: Substitute Completed Squares into the Equation

Substitute the completed squares back into the original equation. We have:\((x+2)^2 - 4 + (y-1)^2 - 1 + 5 = 0\).

Step 5: Simplify the Equation to Find r^2

Combine and simplify the constants:\((x+2)^2 + (y-1)^2 - 4 - 1 + 5 = 0\).This simplifies to:\((x+2)^2 + (y-1)^2 = 0\).Since \((x+2)^2 + (y-1)^2 = 0\), it means \(r^2 = 0\), indicating \(r = 0\).

Step 6: Determine the Center and Radius

From the equation \((x+2)^2 + (y-1)^2 = 0\), the center \((h, k)\) is \((-2, 1)\) and the radius \(r\) is 0.

Key Concepts

Completing the Square

When working with quadratic equations, especially when preparing them for geometric interpretation, completing the square is a vital technique. Imagine you have an equation that includes terms like \(x^2 + 4x\) and \(y^2 - 2y\). These reflect parts of a circle, but they need to be transformed into a recognizable form for easy analysis.

To complete the square, follow these simple steps:

• For the \(x\)-terms \(x^2 + 4x\), take the coefficient of \(x\), which is 4, halve it to get 2, square it to give 4, then add and subtract this number in the equation.
• This transforms \(x^2 + 4x + 4 - 4\) into \((x+2)^2 - 4\).
• Similarly, handle the \(y\)-terms \(y^2 - 2y\) by taking half of -2, which is -1, squaring to get 1, and then adding and subtracting 1.
• The transformation results in \(y^2 - 2y + 1 - 1\) becoming \((y-1)^2 - 1\).

This process is crucial as it allows us to rewrite the equation into a standard form of a circle's equation, making it easy to identify the center and radius.

Circle Center

The center of a circle in Cartesian coordinates is most easily identified once the equation is in its standard form: \((x-h)^2 + (y-k)^2 = r^2\). Here, \((h,k)\) acts as the coordinates for the circle's center.

Through completing the square process in our original equation \(x^2 + y^2 + 4x - 2y + 5 = 0\), we transform it into \((x+2)^2 + (y-1)^2 = 0\).

• Here, the form indicates that the center is at \((h,k) = (-2,1)\).
• It's worth noting that changes in the original terms influence the position of this center. By completing the square, you accurately pinpoint where the circle resides in the plane.

Recognizing the center allows anyone studying the nature of graphs to appreciate how transformations affect shapes.

Circle Radius

The radius of a circle, denoted by \(r\), is the distance from its center to any point on the circle. It offers insight into how "big" the circle is in the plane.

For a given equation like \((x+2)^2 + (y-1)^2 = 0\), it can be deduced that \(r^2\) equates to the constant on the right side. Here, \(r^2 = 0\), meaning \(r = 0\).

• This indicates that the circle is actually a point, with no area, as \(r = 0\).
• While unusual, this emphasizes checking your constants from the transformed equation, as they guide the calculation for radius.

A circle with a zero radius exemplifies a particular instance where the original terms perfectly balance out, leading the geometrical object to collapse into a singular point on the plane.