Problem 39 - Precalculus

Question

Specify the center and radius of each circle. Also, determine whether the given point lies on the circle. $$(x-1)^{2}+(y-5)^{2}=169:(6,-7)$$

Step-by-Step Solution

Answer

Center: (1, 5); Radius: 13; The point (6, -7) lies on the circle.

Step 1: Identify the Circle's Equation Format

The given equation \((x-1)^{2}+(y-5)^{2}=169\) is in the standard form of a circle's equation, which is \((x-h)^{2} + (y-k)^{2} = r^{2}\), with \(h\) and \(k\) representing the center of the circle, and \(r\) as the radius.

Step 2: Determine the Center of the Circle

From the equation, \((x-1)^{2}+(y-5)^{2}=169\), we identify \(h = 1\) and \(k = 5\). Thus, the center of the circle is the point \((1, 5)\).

Step 3: Calculate the Radius of the Circle

The equation shows us that \(r^{2} = 169\). Taking the square root of both sides, we find \(r = 13\). Thus, the radius of the circle is 13 units.

Step 4: Check if the Given Point Lies on the Circle

To check if the point \((6, -7)\) lies on the circle, substitute \(x = 6\) and \(y = -7\) into the equation: \((6-1)^2 + (-7-5)^2 = 169\). Calculate this as \(5^2 + (-12)^2 = 25 + 144 = 169\). Since the left-hand side equals the right-hand side, the point \((6, -7)\) does lie on the circle.

Key Concepts

Standard Form of a Circle

The standard form of a circle’s equation is an easy and powerful tool for understanding the core properties of a circle. This equation is written as \[(x-h)^{2} + (y-k)^{2} = r^{2},\]where:

• \(h\) and \(k\) are the coordinates of the circle’s center,\( (h, k)\).
• \(r\) represents the radius of the circle, which is the distance from the center to any point on the circle.

It's essential to memorize this form because it reveals the circle's central properties at a glance. This makes it simple to identify where the circle is located on a graph and how large it is.
Moreover, this form is helpful in geometry and algebra problems involving circles, such as determining intersections or solving for unknown variables.

Center and Radius

Finding the center and radius of a circle is straightforward when given its equation in standard form. Let's delve into how this can be done using the equation \[(x-1)^{2}+(y-5)^{2}=169.\]
To find the center, we look at the \((x-h)\) and \((y-k)\) parts:

• The term \((x-1)\) implies that \(h = 1\).
• The term \((y-5)\) indicates that \(k = 5\).

Thus, the center of the circle is at the point \((1, 5)\).
Next, to find the radius, we examine the right side of the equation: \(r^{2} = 169\).
By taking the square root of 169, we discover that the radius \(r\) equals 13. This implies that the circle expands 13 units away from the center in every direction, forming a perfect round shape on the plane.

Point on a Circle

Determining whether a specific point resides on a circle involves verifying if it satisfies the circle’s equation. Let’s assess the point \((6, -7)\) in context with our circle equation\((x-1)^{2}+(y-5)^{2}=169.\)
We start by substituting this point into the circle's equation:

• Replace \(x\) with 6 and \(y\) with -7, resulting in:\((6-1)^2 + (-7-5)^2.\)
• Simplify the expression:\((5)^{2} + (-12)^{2} = 25 + 144 = 169.\)

Since both sides of the equation are equal (\(169 = 169\)), this tells us that the point \((6, -7)\) indeed lies on the circle.
Verifying points like this is incredibly useful when plotting graphs or determining the intersection between different shapes or lines.