Problem 8 - Algebra 2 and Trigonometry

Question

In \(3-10\) , the coordinates of point \(P\) on the circle with center at \(C\) are given. Write an equation of each circle: a. in center-radius form b. in standard form. $$ P(0,-3), C(6,5) $$

Step-by-Step Solution

Answer

The circle's equation is \((x-6)^2 + (y-5)^2 = 100\) in both forms.

Step 1: Identify Key Components

For the equation of a circle, we need the center and the radius. Here, the center \( C \) is at \((6, 5)\) and a point on the circle \( P \) is at \((0, -3)\).

Step 2: Find the Radius

The radius \( r \) is the distance between the center \( C(6,5) \) and point \( P(0,-3) \). Using the distance formula, \( r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Here, \( x_1 = 6, y_1 = 5, x_2 = 0, y_2 = -3 \). Calculate \( r = \sqrt{(0-6)^2 + (-3-5)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \).

Step 3: Write the Equation in Center-Radius Form

The center-radius form of the circle's equation is \( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h, k) \) is the center and \( r \) is the radius. Plug in the values, \( (x-6)^2 + (y-5)^2 = 10^2 \), so it becomes \( (x-6)^2 + (y-5)^2 = 100 \).

Step 4: Convert to Standard Form

The standard form is essentially the same as the center-radius form. So write: \( (x-6)^2 + (y-5)^2 = 100 \).

Key Concepts

Distance Formula

The Distance Formula is a way to find the length between two points in a coordinate plane. It's extremely handy when you know the coordinates of two points and you wish to calculate the space separating them. The formula is expressed as:

• \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Here, \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points.
In the context of a circle, if you have a point on the circle and the center of the circle, the Distance Formula helps in determining the radius.
For instance, in the exercise, using the Distance Formula between the center \( C(6,5) \) and the point \( P(0,-3) \), you find the radius is 10 units.

Center-Radius Form

The Center-Radius Form of a circle's equation is a very intuitive way of expressing a circle in algebra.
This form highlights the most critical components: the center of the circle and its radius.

• The general structure is \( (x-h)^2 + (y-k)^2 = r^2 \)
• \((h, k)\) represents the center of the circle
• \(r\) stands for the radius

Using the given example where the center is \( (6, 5) \) and the radius is 10, the equation becomes \((x-6)^2 + (y-5)^2 = 100\).
Writing an equation in this form makes it easy to visually interpret both the position and size of the circle.

Standard Form of Circle Equation

The Standard Form of a circle's equation may look similar to the Center-Radius Form. The two forms are indeed often the same when it comes to circles. It generally looks like this:

• \( (x-h)^2 + (y-k)^2 = r^2 \)

While this might seem repetitive, the importance lies in how equations in Standard Form are often utilized in algebra and geometry for easy manipulation and graphing.
This representation allows for straightforward identification and verification of a circle's center and radius at a glance.
In this exercise, the Standard Form equation \((x-6)^2 + (y-5)^2 = 100\) succinctly describes the same circle, emphasizing the central-coordinates approach which is often required for more advanced algebraic operations.