Problem 54 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 47-56: Find the center and radius of the circle with the given equation. $$ x^{2}+y^{2}-6 x+4 y+13=0 $$

Step-by-Step Solution

Answer

The center is (3, -2) and the radius is 0.

Step 1: Rearrange the Equation

Start by rearranging the given equation to group the terms by variable. The equation is:\[ x^2 + y^2 - 6x + 4y + 13 = 0. \] Group the \(x\) terms together and \(y\) terms together:\[ (x^2 - 6x) + (y^2 + 4y) + 13 = 0. \]

Step 2: Complete the Square for x

To complete the square for \(x\), take the coefficient of \(x\), which is -6, divide it by 2 to get -3, and then square it to get 9. Add and subtract 9 inside the equation:\[ (x^2 - 6x + 9) - 9. \]Rewrite it as a perfect square:\[ (x - 3)^2 - 9. \]

Step 3: Complete the Square for y

To complete the square for \(y\), take the coefficient of \(y\), which is 4, divide it by 2 to get 2, and then square it to get 4. Add and subtract 4 inside the equation:\[ (y^2 + 4y + 4) - 4. \]Rewrite it as a perfect square:\[ (y + 2)^2 - 4. \]

Step 4: Simplify the Equation

Substitute the completed squares back into the original equation:\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 + 13 = 0. \]Simplify the equation:\[ (x - 3)^2 + (y + 2)^2 = 0. \] \[ -9 - 4 + 13 = 0. \]Combine the constant terms, we get 0.

Step 5: Determine the Center and Radius

The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. From the equation \((x - 3)^2 + (y + 2)^2 = 0\), we see the center is \((3, -2)\) and \(r^2 = 0\). Therefore, the radius \(r = 0\).

Key Concepts

Center of a Circle

The center of a circle in the coordinate plane is a crucial concept that can be easily understood once you grasp the basics of the circle equation. For a circle represented by the equation \((x - h)^2 + (y - k)^2 = r^2\), the terms \(h\) and \(k\) are of utmost importance. They define the coordinates of the center of the circle. In this standard form, \((h, k)\) gives you the exact point around which the circle is perfectly balanced.
Let's break it down. In the given exercise, after rearranging and completing the square, the equation turns into \((x - 3)^2 + (y + 2)^2 = 0\). In this equation, note the expressions inside the parentheses are offset from zero by the values \(h = 3\) and \(k = -2\).
Therefore, the center of the circle is at the point \((3, -2)\). This means if you were to plot this circle, the center would be located precisely at the coordinate point \((3, -2)\) on the Cartesian plane. Understanding the center helps in analyzing the circle's positioning related to other geometrical shapes and objects.

Radius of a Circle

The radius of a circle is another fundamental aspect that defines its size. By understanding what the radius represents and how it can be determined from the circle equation, you'll gain a clearer insight into circle geometry. The general equation \((x - h)^2 + (y - k)^2 = r^2\) shows that \(r\) is the circle's radius.
In essence, the radius is the constant distance from the center to any point on the circle's edge. Now, looking at the transformed equation from our exercise, we ended up with \((x - 3)^2 + (y + 2)^2 = 0\). Here, notice that \(r^2 = 0\), indicating that \(r = 0\).
This unusual result suggests a rather special case: rather than forming a standard circle, the collection of points effectively all coincide at the center. In technical terms, this is often referred to as a "degenerate circle" or simply a point, since the circle has no extent beyond its center point.

Completing the Square

Completing the square is a powerful method used to transform quadratic expressions into perfect square forms, revealing valuable geometric insights about equations, particularly those involving circles. This technique helps convert a circle's general equation into the standard form, making the center and radius apparent.
To complete the square, focus on the quadratic portions of the variables. First, manipulate the variable terms (such as \(x^2 - 6x\)) to form a perfect square. This involves taking half of the coefficient of the linear term \(-6\), resulting in \(-3\), squaring it to get \(9\), and adding/subtracting \(9\) accordingly. As a result, \(x^2 - 6x\) becomes \((x - 3)^2 - 9\).
Similarly, apply the method to the \(y\) terms: convert \(y^2 + 4y\) to \((y + 2)^2 - 4\) by following the same steps: halve \(4\) to obtain \(2\), square it to get \(4\).
These transformations simplify the equation into the neat form \((x - 3)^2 + (y + 2)^2 = 0\). Completing the square not only simplifies calculations but also offers a visual representation of algebraic expressions, turning them into familiar, easy-to-interpret geometrical forms.