Problem 11 - Calculus Single Variable

Question

The Cartesian equation of a circle is given. Sketch the circle and specify its center and radius. \(x^{2}+y^{2}-y=0\)

Step-by-Step Solution

Answer

The center of the circle is \((0, \frac{1}{2})\) and the radius is \(\frac{1}{2}\).

Step 1: Identify the Circle Equation Format

Start with the given equation of the circle: \(x^{2}+y^{2}-y=0\). This equation is not in the standard form. 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.

Step 2: Rearrange and Complete the Square for y

Rewrite the equation to focus on \(y\): \(x^{2}+y^{2}-y=0\) becomes \(x^{2} + (y^{2} - y) = 0\). To complete the square for \(y\), take half of the coefficient of \(y\), which is \(-\frac{1}{2}\), square it to get \(\left(-\frac{1}{2}\right)^2 = \frac{1}{4}\). Add and subtract \(\frac{1}{4}\) inside the equation: \(x^{2} + (y^{2} - y + \frac{1}{4} - \frac{1}{4}) = 0\).

Step 3: Simplify the Completed Square

The expression \((y^{2} - y + \frac{1}{4})\) becomes \((y - \frac{1}{2})^{2}\). Substitute back into the equation: \(x^{2} + (y - \frac{1}{2})^{2} - \frac{1}{4} = 0\).

Step 4: Rewrite in Standard Circle Form

Add \(\frac{1}{4}\) to both sides to get: \(x^{2} + (y - \frac{1}{2})^{2} = \frac{1}{4}\). Now, the equation is in the standard form \((x-h)^2 + (y-k)^2 = r^2\), where \(h = 0\), \(k = \frac{1}{2}\), and \(r^2 = \frac{1}{4}\).

Step 5: Identify the Center and Radius

From the equation \(x^2 + (y - \frac{1}{2})^2 = (\frac{1}{2})^2\), the center is \((0, \frac{1}{2})\) and the radius is \(\frac{1}{2}\).

Step 6: Sketch the Circle

To sketch, plot the center of the circle at \((0, \frac{1}{2})\) on the Cartesian plane. From this point, mark a circle with a radius of \(\frac{1}{2}\), extending evenly in all directions (up, down, left, and right) from the center point.

Key Concepts

Completing the Square

Completing the square is a technique used to transform quadratic equations into a form that reveals important properties about the graph. When we talk about circles, this process helps convert the given equation into a more useful format. The general idea is to manipulate an equation to have perfect square terms, making it easier to deduce the center and radius of the circle.

In our example, the original equation is:

• \(x^2 + y^2 - y = 0\)

We focus on the quadratic part of the equation for the variable \(y\). By identifying the coefficient of \(y\), which is \(-1\), we take half of this coefficient, square it, and then add and subtract this value to complete the square. This value is obtained as:

• \(\left(-\frac{1}{2}\right)^2 = \frac{1}{4}\)

By completing the square, we transform part of our equation into a form that can easily be written as a squared binomial, which is key for rewriting the equation in standard form.

Standard Form of Circle Equation

The standard form of a circle's equation is given by:

• \((x-h)^2 + (y-k)^2 = r^2\)

In this form, \( (h, k) \) represents the center of the circle, and \( r \) denotes its radius. This form makes it simple to identify the circle's features just by looking at the equation.

Once we've completed the square for the \(y\) part of our equation, we can rewrite the equation as:

• \( x^2 + (y-\frac{1}{2})^2 = \frac{1}{4} \)

This matches the standard form with \( h = 0 \), \( k = \frac{1}{2} \), and \( r^2 = \frac{1}{4} \). Converting an equation into this form aids in quickly pinpointing the exact position and size of the circle.

Center of a Circle

The center of a circle provides a reference point from which all points on the circle are equidistant. It is usually represented as a coordinate pair \((h, k)\). From the standard equation of the form \((x-h)^2 + (y-k)^2 = r^2\), the values \(h\) and \(k\) emerge directly as the center.

In our example, after the equation has been rewritten to standard form, we identify:

• \( h = 0 \)
• \( k = \frac{1}{2} \)

This means the center of the circle is located at \((0, \frac{1}{2})\) on the Cartesian coordinate plane. Knowing the center is crucial for sketching the circle and understanding its geometric relationship to other objects in the plane.

Radius of a Circle

The radius of a circle is the distance from the center of the circle to any point on its perimeter. It is a fundamental property that directly affects the circle's size. In the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\), \(r\) represents the circle's radius.

Our example, when transformed to its standard form, gives:

• \((x-0)^2 + (y-\frac{1}{2})^2 = \left(\frac{1}{2}\right)^2 \)

The equation explicitly states \(r^2 = \frac{1}{4}\). By taking the square root of both sides, we find the radius \(r = \frac{1}{2}\). This short distance from center to perimeter highlights how compact the circle is. The radius sets the scale of the circle when sketching and determining its area.