Problem 25 - Intermediate Algebra

Question

Write the equation of a circle in standard form with the following properties. Center at \((6,8) ;\) radius 5

Step-by-Step Solution

Answer

The standard form of the circle is \((x - 6)^2 + (y - 8)^2 = 25\).

Step 1: Identify the Center Coordinates

The center of the circle is given as \((6, 8)\). That means the coordinates \((h, k)\) in the standard equation of a circle\( (x-h)^2 + (y-k)^2 = r^2\) are \(h = 6\) and \(k = 8\).

Step 2: Use the Given Radius Value

The radius of the circle is given as \(5\). Therefore, \(r^2\) is equal to \(25\) because \(5^2 = 25\).

Step 3: Write the Standard Circle Equation

Substitute the values of \(h\), \(k\), and \(r^2\) into the standard circle equation:\[(x - 6)^2 + (y - 8)^2 = 25\]

Key Concepts

Standard Form

The standard form of a circle's equation is a way to easily describe all the points that make up the circle. It's given by the formula:

• \( (x - h)^2 + (y - k)^2 = r^2 \)

In this formula:

• \((x, y)\) are the coordinates of any point on the circle.
• \((h, k)\) is the center of the circle.
• \(r\) is the radius of the circle.

This equation helps us see how each point on the circle relates to its center \((h, k)\). When a point is on this circle, it will always satisfy this equation. That means when you plug the coordinates \((x, y)\) of any point on the circle into the equation, the equation holds true, thus making the calculations very straightforward.

Radius of a Circle

The radius is a crucial part of understanding circles. The radius of a circle is the straight-line distance from its center to any point on the circle's perimeter.

For instance, in our given exercise, the radius is \(5\). But why is it squared in the standard form equation? It's because the formula uses the circle's equation derived from the Pythagorean theorem, which involves squares.

• When you know the radius, you can easily find \(r^2\) (radius squared), which is used in the equation.

So, if the radius is \(5\), then \(r^2 = 25\). This is a very common step when dealing with any circle equation.

Center of a Circle

The center of a circle, marked as \((h, k)\) in the standard form equation, is a fixed point that is equidistant from every point on the circle's perimeter. In simpler terms, if you start at the center and go outwards in any direction, the path will always be the length of the radius.

• For our problem, the center is given at the point \((6, 8)\).

Understanding the center is key to writing the circle's equation correctly. You simply plug the \(h\) and \(k\) values directly into the standard form equation. This ensures that the formula correctly represents every point that is a radius's distance from \((h, k)\), thus forming the circle. By knowing the center, you have a precise starting point for constructing or visualizing the circle.