Problem 2 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 4, find an equation of the circle with center at \(C\) and radius \(r\). Write the equation in both the centerradius form and the general form. $$ C(0,0), r=8 $$

Step-by-Step Solution

Answer

The circle's center-radius form is \(x^2 + y^2 = 64\); the general form is \(x^2 + y^2 - 64 = 0\).

Step 1: Understand the Circle's Equation in Center-Radius Form

For a circle with center \(C(h, k)\) and radius \(r\), the equation in center-radius form is \[ (x - h)^2 + (y - k)^2 = r^2 \].

Step 2: Substitute the Given Values

Given that the center \(C\) is \( (0,0) \) and the radius \( r \) is \( 8 \), substitute these values into the center-radius form equation to get \[ (x - 0)^2 + (y - 0)^2 = 8^2 \].

Step 3: Simplify the Equation

Simplify the equation by removing the zeros and squaring the radius: \[ x^2 + y^2 = 64 \].

Step 4: Write the General Form Equation of the Circle

To write the general form of the equation of the circle, convert the simplified center-radius form \[ x^2 + y^2 = 64 \] to the general form, which is \[ Ax^2 + Ay^2 + Bx + Cy + D = 0 \]. Here, \[ A = 1 \], \[ B = 0 \], \[ C = 0 \], and \[ D = -64 \]. Therefore, the equation is: \[ x^2 + y^2 + 0x + 0y - 64 = 0 \].

Key Concepts

center-radius form

The center-radius form of an equation of a circle is a popular way to represent circles in mathematics. It precisely defines the position and size of the circle based on its center coordinate and radius length. The standard format for this form is: \[ (x - h)^2 + (y - k)^2 = r^2 \] Here:

• \textbf{(h, k)} is the center of the circle.
• The distance from the center to any point on the circle is \textbf{r} (the radius).

For example, if the center is at (0, 0) and the radius is 8, you simply plug in these values to get the equation: \[ (x - 0)^2 + (y - 0)^2 = 8^2 \] Simplifying the zeros and squaring the radius gives: \[ x^2 + y^2 = 64 \] This makes it clear where the center is and how large the circle is.

general form

While the center-radius form is very intuitive, sometimes it is necessary to convert the equation into the general form. This is often required in higher-level mathematics and various applications, such as problem-solving and graphing software.The general form of a circle's equation is: \[ Ax^2 + Ay^2 + Bx + Cy + D = 0 \] Let's take our simplified center-radius form \[ x^2 + y^2 = 64 \] and rewrite it into the general form:- Notice that this equation already covers the \textbf{Ax}² and \textbf{Ay}² terms.- For terms \textbf{Bx} and \textbf{Cy}, both are zero because there are no x or y terms in the simplified equation.- Finally, move the constant term (64) to the other side to set the equation to 0.Converting, we get: \[ x^2 + y^2 + 0x + 0y - 64 = 0 \] So, in this specific case: \[ A = 1 \], \[ B = 0 \], \[ C = 0 \], and \[ D = -64 \].

substitution

Substitution is the action of inserting values into an equation. In problems involving the equation of a circle, you often use substitution to plug in the coordinates of the center (\textbf{h}, \textbf{k}) and the radius (\textbf{r}).Starting with the center-radius form: \[ (x - h)^2 + (y - k)^2 = r^2 \]If given a center \textbf{C(0, 0)} and a radius \textbf{r = 8}, you substitute these values:

• Replace \textbf{h} with \textbf{0}
• Replace \textbf{k} with \textbf{0}
• Replace \textbf{r} with \textbf{8}

This substitution changes the equation to:\[ (x - 0)^2 + (y - 0)^2 = 8^2 \]Simplify it:\[ x^2 + y^2 = 64 \]Hence, through substitution, you can easily find the circle's equation with specific center and radius values.