Problem 62 - Precalculus Essentials

Question

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+x+y-\frac{1}{2}=0$$

Step-by-Step Solution

Answer

The standard form of the given equation is \(\left(x+\frac{1}{2}\right)^{2} + \left(y+\frac{1}{2}\right)^{2} = 1\), the center of the circle is (-0.5, -0.5), and radius is 1.

Step 1: Rewriting Given Equation

Regroup the equation so that all the \(x\)'s and \(y\)'s variables are together. Take the constant to the opposite side of the equation, yielding: \((x^{2}+x) + (y^{2}+y) = \frac{1}{2}\)

Step 2: Complete The Square For X And Y

Next, use the completing the square method on the \(x\) and \(y\) terms. For any quadratic in the form \(ax^{2} + bx\), \((x+\frac{b}{2a})^{2} = x^{2} + bx + \frac{b^{2}}{4a^{2}}\). Apply that to x and y in the equation separately, giving the equation: \(\left(x+\frac{1}{2}\right)^{2} + \left(y+\frac{1}{2}\right)^{2} = \frac{1}{4} + \frac{1}{4} + \frac{1}{2}\)

Step 3: Simplify The Equation To Find Standard Form

Reduce the equation to the simplest form which gives: \(\left(x+\frac{1}{2}\right)^{2} + \left(y+\frac{1}{2}\right)^{2} = 1\)

Step 4: Determine The Center And Radius

The equation is now in the standard form, which is \((x - h)^2 + (y - k)^2 = r^2\). From here, we see that \(h = -\frac{1}{2}\), \(k = -\frac{1}{2}\) and \(r = 1\). So, the center of the circle is \((-0.5, -0.5)\) and the radius is 1.

Step 5: Graph The Equation

Finally, the equation can be graphed on an xy-plane. The center of the circle is at (-0.5, -0.5). Draw a circle with a radius of 1 unit around this point.

Key Concepts

standard form of a circle

The standard form of a circle's equation is essential in geometry. It makes understanding and graphing circles simple.
To put an equation into standard form, we use the format:

• \((x - h)^2 + (y - k)^2 = r^2\)

This structure clearly organizes the circle's properties.
Here, \( (h, k) \) represents the circle's center, while \( r \) is its radius.
By completing the square, terms are rearranged to fit this tidy structure, enabling easy identification of the circle's main elements.

center of a circle

The center of a circle is crucial as it defines the circle’s position on an xy-plane.
In the standard form equation, \((x - h)^2 + (y - k)^2 = r^2\), the coordinates \((h, k)\) are the center.
For example, in the equation

• \((x + \frac{1}{2})^2 + (y + \frac{1}{2})^2 = 1\)

the center is

• \((-\frac{1}{2}, -\frac{1}{2})\)

This point is derived from reversing the signs within the parentheses to find \(h\) and \(k\).
Visualizing this point helps in drawing the precise location of the circle on the graph.

radius of a circle

The radius is the distance from the center to any point on the circle. It’s a key aspect that determines the circle’s size.
In the standard form equation,

• \(r^2\) appears on the right side.

To find the radius, simply take the square root of this term.
For example, in the equation

• \((x + \frac{1}{2})^2 + (y + \frac{1}{2})^2 = 1\)

we have

• \(r^2 = 1\), so \(r = \sqrt{1} = 1\).

This means the circle has a uniform distance of 1 unit from the center to its edge.