Problem 5 - 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. $$ P(4,0), C(0,0) $$

Step-by-Step Solution

Answer

The equation of the circle is \(x^2 + y^2 = 16\).

Step 1: Identify the Center and Radius

The problem states that the circle's center is at point \( C(0,0) \) and point \( P(4,0) \) is on the circle. The radius \( r \) of the circle can be found by calculating the distance from the center \( C \) to point \( P \). Since \( P \) is on the circle, the distance is equal to the radius.

Step 2: Calculate the Radius

Use the distance formula to calculate the radius \( r \): \[ r = \sqrt{(4-0)^2 + (0-0)^2} = \sqrt{16} = 4 \]. Thus, the radius of the circle is 4.

Step 3: Write Equation in Center-Radius Form

The center-radius form of a circle's equation is \[ (x-h)^2 + (y-k)^2 = r^2 \] where \((h, k)\) is the center of the circle and \( r \) is the radius. Substitute \( h = 0, k = 0, r = 4 \) into the equation: \[ (x-0)^2 + (y-0)^2 = 4^2 \].

Step 4: Simplify Center-Radius Form

Simplify the equation from Step 3: \[ x^2 + y^2 = 16 \]. This is the center-radius form of the circle's equation.

Step 5: Write Equation in Standard Form

The standard form of the circle's equation is the same as the simplified center-radius form for this particular circle: \[ x^2 + y^2 = 16 \]. Thus, the standard form is the same as the center-radius form for a circle centered at the origin.

Key Concepts

Center-Radius Form

To talk about circle equations, you first need to understand the center-radius form. This way of expressing a circle's equation centers around the point from which you measure the radius. When a circle is situated with its center at a point \(h,k\), and the radius is \(r\), the equation looks like this: \((x-h)^2 + (y-k)^2 = r^2\).
This form is very intuitive for circles:

• \(h, k\) are the coordinates of the center of the circle.
• \(r\) is the radius of the circle.

For example, if the center of the circle is at the origin \(0, 0\) and the radius is \(4\), the equation becomes \( (x-0)^2 + (y-0)^2 = 4^2\).
When you simplify, it's easy to see this equation become \((x^2 + y^2 = 16)\). This equation right here tells you everything about the size and position of the circle.

Standard Form

Interestingly, when your circle's center is at the origin, the center-radius form does not need alteration to become the standard form. These two forms are identical in such cases because the simplifications result in the same expression.
For circles not centered at the origin, the standard form offers a more generalized way to compare and manipulate all sorts of equations.

• Standard form highlights \(x^2 + y^2\) terms exactly as they are.

In this exercise, our circle's equation is \(x^2 + y^2 = 16\) for both forms. It underscores the easy math involved when circles are centered at the origin.

Distance Formula

To find a circle's radius, the distance formula is your go-to tool. This formula computes the distance between two points, crucial when establishing the radius from the circle's center to any point on the circle.
The distance formula looks like this:

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

For our example, the circle's center is at \(C(0,0)\) and a point on the circle is \(P(4,0)\):

• Plug into the formula: \(d = \sqrt{(4 - 0)^2 + (0 - 0)^2} = 4 \).

Therefore, the distance or radius is \(4\). It simplifies the process and provides precision in radius calculation.

Radius Calculation

Let's take a moment to focus in on radius calculation using all we know so far. For a circle centered at point \(C(h, k)\), the radius is the distance from \(C\) to any point \(P(x, y)\) on the circle.

• The distance formula, \( r = \sqrt{(x-h)^2 + (y-k)^2} \), gives us the handy tool for this calculation.

Since our circle's center is \(C(0, 0)\) and \(P(4, 0)\) is on the circle:

• We calculate: \( r = \sqrt{(4-0)^2 + (0-0)^2} = 4 \).

This calculation is straightforward yet critical, as it sets up our entire equation for both the center-radius and standard forms.
Don’t forget – knowing the radius is the key to unlocking any circle's size and corresponding equations.