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Chapter 13: Problem 72

Write each equation in standard form, if it is not alreacty so, and graph it. If the graph is a circle, give the coordinates of its center and its radius. If the graph is a parabola, give the coordinates of its vertex. $$ (x-3)^{2}+(y-2)^{2}=36 $$

Short Answer

Expert verified
The graph is a circle centered at (3, 2) with a radius of 6.

Step by step solution

01

Identify the Type of Equation

The given equation is \[ (x-3)^2 + (y-2)^2 = 36 \]Notice that it matches the general form of a circle's equation:\[ (x-h)^2 + (y-k)^2 = r^2 \] where \( h \) and \( k \) are the coordinates of the center, and \( r \) is the radius.
02

Rewrite Equation in Standard Form

Since the equation is already in the standard form for a circle, no further rearrangement is necessary. We have\[ (x-3)^2 + (y-2)^2 = 36 \] in its standard form.
03

Identify the Center of the Circle

In the equation \( (x-3)^2 + (y-2)^2 = 36 \), compare it with \( (x-h)^2 + (y-k)^2 = r^2 \). The center \((h, k)\) is \((3, 2)\).
04

Determine the Radius

The square of the radius \( r^2 \) is 36, so the radius \( r \) is the square root of 36. Therefore, \( r = 6 \).
05

Graph the Circle

Draw a graph with a center at the point \((3, 2)\). From this center point, measure a distance of 6 units in all directions (up, down, left, and right) to outline the circle. Connect these points smoothly to form the circle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Standard Form of a Circle Equation
A circle equation in its standard form looks like \[ (x-h)^2 + (y-k)^2 = r^2 \] This specific structure gives us valuable information about the circle it represents. The variables \( h \) and \( k \) in the equation are not arbitrarily chosen鈥攖hey denote the coordinates of the circle's center.
Meanwhile, the \( r^2 \) represents the square of the circle's radius. This equation format facilitates easy identification of both the circle's center and its radius.
  • The term \((x-h)^2\) indicates how much a point \(x\) deviates from the center's x-coordinate \(h\).
  • Similarly, \((y-k)^2\) shows the deviation of \(y\) from the center's y-coordinate \(k\).
  • The addition of these squared terms ensures all points on the circle share the same total deviation, defined by the value \(r^2\).
This equation's primary use is in graphing and calculating other properties of circles.
Finding the Center Coordinates of the Circle
To locate the center of a circle on a graph, look at the equation in standard form.
The center's coordinates, \( (h, k) \), are obtained directly from the expressions \((x-h)^2\) and \((y-k)^2\).
For the equation \[ (x-3)^2 + (y-2)^2 = 36, \] the center is clearly at point \( (3, 2) \):
  • The term \((x-3)^2\) suggests the x-coordinate of the center is 3.
  • The component \((y-2)^2\) indicates the y-coordinate of the center is 2.
Thus, the center of the circle is positioned at \( (3, 2) \), allowing you to easily underscore the circle's location on a plane.
Calculating the Radius from the Circle Equation
Finding the radius of a circle from its equation is straightforward once you understand the standard form.
The term \( r^2 \) is present on the right side of the equation, and taking its square root reveals the radius \( r \).
In the equation \[ (x-3)^2 + (y-2)^2 = 36, \] the value \( r^2 \) equals 36.Here鈥檚 how you calculate the radius:
  • Identify \( r^2 \), which in this case is 36.
  • Find the square root of 36 to determine \( r \).
  • Thus, \( r = \sqrt{36} = 6 \).
Understanding how to extract the radius from the equation helps you graph the circle accurately and understand its size.

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