Problem 51 - Algebra and Trigonometry

Question

\(49-54\) . Find the center and radius of the circle, and sketch its graph. $$ (x-3)^{2}+y^{2}=16 $$

Step-by-Step Solution

Answer

Center: (3, 0), Radius: 4. Sketch is a circle at (3, 0) with radius 4.

Step 1: Identify the Equation Form

The given equation \[ (x-3)^2 + y^2 = 16 \]is in the standard form of a circle equation: \[ (x-h)^2 + (y-k)^2 = r^2 \]where \((h, k)\) is the center and \(r\) is the radius.

Step 2: Determine the Center of the Circle

From the equation, compare with the standard form: \((x-3)^2 + (y-0)^2 = 16\).Here, the center \((h, k)\) is \((3, 0)\).

Step 3: Calculate the Radius of the Circle

Set \(r^2 = 16\). To find \(r\), take the square root of both sides:\[ r = \sqrt{16} = 4 \].So, the radius of the circle is 4.

Step 4: Sketch the Graph

Plot the center of the circle at the point \((3, 0)\) on a coordinate plane. Use the radius of 4 to draw the circle around this center. The circle should pass through the points 4 units away from the center in all directions.

Key Concepts

Standard Form of a Circle

Understanding the standard form of a circle's equation is crucial for analyzing its features. The general equation is \((x-h)^2 + (y-k)^2 = r^2\), where:

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

This equation clearly shows the relationship between the coordinates of any point on the circle and its center.
It tells us how far each point is from the center using the constant radius \(r\). In this setup, all points \((x, y)\) satisfy the equation, meaning they lie precisely on the circle's edge.
One of the primary benefits of using the standard form is how it makes identifying the center and radius straightforward, which saves time during calculations and graphing.

Center of a Circle

The center of a circle is a key feature and is heavily embedded in its equation. From the standard form \((x-h)^2 + (y-k)^2 = r^2\), the center is \((h, k)\).
These \(h\) and \(k\) values are taken directly from the equation. When you compare the equation \((x - 3)^2 + y^2 = 16\) to the standard form:

• \((x-h)^2 = (x-3)^2\) gives \(h = 3\).
• \((y-k)^2 = y^2\) implies \(k = 0\).

Thus, the center of this particular circle is \((3, 0)\). This understanding helps you accurately position the circle on a graph, ensuring that its shape is correct in a Cartesian coordinate system.

Radius of a Circle

The radius of a circle defines how large the circle is, and it is derived from the equation in a straightforward manner. In the standard form \((x-h)^2 + (y-k)^2 = r^2\), \(r^2\) is set equal to a constant.
In the provided equation \((x-3)^2 + y^2 = 16\), we identify this constant as 16, meaning:

• \(r^2 = 16\).

To find the radius \(r\), simply take the square root of this value:

• \(r = \sqrt{16} = 4\).

This shows that the radius is 4. The radius determines how far the edge of the circle is from the center in all directions,
allowing us to draw the circle accurately by counting four units outwards in any direction from the center on a graph.