Problem 26 - Algebra 2

Question

For each equation, find the center and radius of the circle. $$ (x-3)^{2}+(y-7)^{2}=96 $$

Step-by-Step Solution

Answer

The center of the circle is at \( (3, 7) \), and its approximate radius is 9.80.

Step 1: Identify the Center

To find the center we need to identify the values of \(h\) and \(k\) in our equation. In the equation \((x-3)^{2}+(y-7)^{2}=96\), \(h\) is the number subtracted from \(x\) inside the braces, and \(k\) is the number subtracted from \(y\) inside the braces. So, our center \((h, k)\) is \((3, 7)\).

Step 2: Identify the Radius

To find the radius, we look at the right side of the equation, which equals \(r^2\). In the equation \((x-3)^{2}+(y-7)^{2}=96\), \(r^{2}\) equals 96. To get \(r\), we need to take the square root of 96, which equals \(\sqrt{96} = 9.80\) (approximately, to two decimal places).

Key Concepts

Circle Center

To understand how to find the center of a circle from its equation, we need to look closely at the equation's structure. An equation of a circle is typically written in the form

• \((x-h)^{2} + (y-k)^{2} = r^{2}\)

Here,

• \((h, k)\) represents the coordinates of the circle's center.
• The values \(h\) and \(k\) are simply the numbers subtracted from \(x\) and \(y\) in the equation.

In the exercise provided, the equation is

• \((x-3)^{2}+(y-7)^{2}=96\)

This means that the center of the circle is

• \((3, 7)\)

because \(3\) is subtracted from \(x\) and \(7\) from \(y\). Remember, the signs for \(h\) and \(k\) in the center are the opposite of those seen in the equation.

Circle Radius

The radius of a circle is a crucial part of understanding its size. In the standard circle equation

• \((x-h)^{2} + (y-k)^{2} = r^{2}\)

the term on the right side, \(r^{2}\), represents the square of the radius. When we need to find the actual radius, \(r\), we simply take the square root of \(r^{2}\).In the given example, the equation

• \((x-3)^{2} + (y-7)^{2} = 96\)

tells us that \(r^{2}\) equals \(96\). To find \(r\), calculate

• \(\sqrt{96} \approx 9.80\)

This means the radius of the circle is approximately \(9.80\). Knowing how to manipulate this part of the equation helps in quickly determining how far any point on the circle is from its center.

Equations of Circles

The equation of a circle in mathematics provides a robust way to represent circular shapes using algebraic expressions. It's a straightforward yet powerful tool. The general 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. This form encapsulates all information needed about the circle's size and location. Focusing on this structure allows us to effortlessly plug in the values for the center and read the circle's radius from the square term on the right side of the equation. Equations like these not only help in theoretical mathematics but also have practical applications. They are used in engineering, computer graphics, and even physics to model and manipulate circular paths or objects. Understanding them is a foundational skill in geometry, offering clarity for more complex topics dealing with curves and graphical data structures.

Standard Form of a Circle Equation

Understanding the standard form of a circle equation is essential for grasping how circles are expressed mathematically. The standard form is typically written as

• \((x-h)^{2} + (y-k)^{2} = r^{2}\),

where \( (h, k) \) corresponds to the center and \( r \) is the radius. This form is clean and direct, making it easy to identify both center and radius at a glance.Why use the standard form? Because it gives you a clear method to interpret the circle's placement and size without additional conversions. This is especially useful when graphing, as it immediately shows which point is central and how far points on the circle are from this center.Whenever tackling problems involving circles, it's crucial to convert any given circle equation into this standard form, if it's not already. By doing so, you are unlocking the equation's full potential, making geometric visualization and problem-solving much simpler and more intuitive. This standard form is a bedrock principle in geometry, paving the way to understanding broader mathematical landscapes, including analytical geometry and calculus.