Problem 10 - Calculus Single Variable

Question

The Cartesian equation of a circle is given. Sketch the circle and specify its center and radius. \((x-3)^{2}+y^{2}+y=1\)

Step-by-Step Solution

Answer

Center: (3, -1/2), Radius: \(\sqrt{5}/2\).

Step 1: Identify Standard Form of Circle Equation

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. Our given equation does not fit this form exactly, so we will need to rearrange terms to match it.

Step 2: Completing the Square for y

The given equation is \((x-3)^2 + y^2 + y = 1\). Focus on the \(y\) terms: \(y^2 + y\). To complete the square, take half of the coefficient of \(y\) (which is 1), square it, and add it inside along with subtracting the same amount: \((y^2 + y + (1/2)^2 - (1/2)^2)\). This gives us \((y + 1/2)^2 - 1/4\).

Step 3: Rewriting the Equation

Now, rewrite the equation by incorporating the completed square: \((x-3)^2 + (y + 1/2)^2 - 1/4 = 1\). To maintain equality, add \(1/4\) to both sides to balance the equation: \((x-3)^2 + (y+1/2)^2 = 1 + 1/4\). This simplifies to \((x-3)^2 + (y+1/2)^2 = 5/4\).

Step 4: Identify Center and Radius

The equation now matches the standard circle form \((x-h)^2 + (y-k)^2 = r^2\). From \((x-3)^2 + (y+1/2)^2 = 5/4\), we identify the center \((h, k)\) of the circle as \((3, -1/2)\) and the radius \(r\) as \(\sqrt{5/4}\) or \(\sqrt{5}/2\).

Step 5: Sketch the Circle

On a coordinate plane, plot the center of the circle at \((3, -1/2)\). Use the radius \(\sqrt{5}/2\) to measure out from the center in all directions to form a circle. Note that \(\sqrt{5}/2\) is approximately 1.118, so the circle extends a little over 1 unit from the center.

Key Concepts

Standard Form of a Circle

When dealing with circles in coordinate geometry, using the standard form of a circle's equation makes it easier to identify important features.
The standard form is written as \((x-h)^2 + (y-k)^2 = r^2\).
In this formula:

• \((h, k)\) represents the circle's center coordinates.
• \(r\) stands for the radius of the circle.

This form allows for quick identification and makes graphing straightforward. However, given equations might not always appear in this form initially, like our example. Hence, rearranging and completing the square are necessary to transform it into the standard form.

Cartesian Equation

The Cartesian equation is used to describe circles using algebraic terms in a plane.
The beauty of Cartesian equations is that they allow you to represent geometric shapes through expressions involving \(x\) and \(y\).
For circles, they often appear as quadratic equations.To work with a circle's Cartesian equation, convert it to the standard form by rearranging and using algebraic techniques such as completing the square.
This transformation simplifies the identification of the circle’s center and radius, enabling easier analysis and sketching.

Center and Radius

The concepts of the center and radius are fundamental in defining and graphing a circle.
The center, \((h, k)\), is the point around which the circle is perfectly symmetrical.- **Finding the Center:** By completing the square in a given quadratic equation, as shown in the original solution, you can identify the \((h, k)\) values directly from the standard form.
In our example, the center was pinpointed at \((3, -1/2)\).
- **Finding the Radius:** The radius is the distance from the center to any point on the circle’s edge.
It is found as \(r = \sqrt{\text{constant term}}\).
In the example, it was revealed to be \(\sqrt{5}/2\), which equals approximately 1.118.Sketching a circle with known center and radius becomes a straightforward task once they are determined. Simply measure from the center according to the radius unit to draw the circle evenly around the central point.