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Chapter 10: Problem 54

Use a graphing calculator to graph each equation. $$ x^{2}+(y-2)^{2}=4 $$

Short Answer

Expert verified
The graph is a circle centered at (0, 2) with a radius of 2.

Step by step solution

01

Identify the Equation Type

The given equation is a circle equation in the standard form, written as \( (x - h)^2 + (y - k)^2 = r^2 \). Here, \( h = 0 \), \( k = 2 \), and \( r^2 = 4 \). Thus, the circle is centered at \( (0, 2) \) with a radius of \( 2 \).
02

Graph the Circle

With the center of the circle at \( (0, 2) \), plot this point on the graphing calculator. Then, draw a circle with radius \( 2 \) around this center point. Ensure that the circle equally extends to the left, right, up, and down from the center point due to its symmetry.
03

Check the Symmetry

Since the circle is centered at \( (0, 2) \) and has a constant radius, it should appear perfectly symmetrical horizontally and vertically about the y-axis line \( y = 2 \). Verify that the graph confirms this symmetry.

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Key Concepts

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

Circle Equation
An equation of a circle is usually presented in its standard form, which is \( (x - h)^2 + (y - k)^2 = r^2 \). This equation is pivotal when dealing with circles as it clearly shows the position of the circle on the Cartesian plane. The components \( h \) and \( k \) indicate the horizontal and vertical shifts from the origin, determining the center of the circle at \((h, k)\). Meanwhile, \( r^2 \) denotes the square of the radius \( r \), offering insight into the size of the circle. Using this equation, you can easily identify the core properties of a circle, which will be useful when graphing or performing other mathematical operations on circles.
Radius and Center of a Circle
Understanding the radius and center of a circle is key in graphing it correctly. The center \((h, k)\) is the fixed point from which the circle is equally distant in all directions. This understanding helps you accurately position the circle on a graph by plotting a single point and using a compass or a graphing calculator to scope an equal distance in every direction.
  • Center: For the equation \( (x - 0)^2 + (y - 2)^2 = 4 \), the center is at \((0, 2)\). This means the circle is positioned 2 units up the y-axis from the origin.
  • Radius: The equation \( r^2 = 4 \) tells us that \( r = \sqrt{4} = 2 \). Therefore, the radius of the circle is 2 units, indicating the distance from the center to any point on the circle.
This clear identification allows for precise and symmetric graphing on any Cartesian plane.
Symmetry of Circles
Symmetry in circles is an essential aspect of their geometry, contributing to their aesthetic appeal and ease of mathematical handling. A circle is symmetrical because all points are at an equal radius from the center, providing uniformity across the shape.
Circles demonstrate several kinds of symmetry:
  • Horizontal Symmetry: A circle remains unchanged when divided horizontally across its center, especially in this case, around the line \( y = 2 \).
  • Vertical Symmetry: Similarly, when split vertically, a circle will mirror itself perfectly.
  • Rotational Symmetry: A circle's symmetry enables it to be rotated around its center at any degree, and it will still appear complete and unchanged.
This inherent symmetry ensures a perfect shape, making circles a wonderful subject for graphing and geometric study.

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Most popular questions from this chapter

Write each equation in standard form, if it is not already so, and graph it. The problems include equations that describe circles, parabolas, ellipses, and hyperbolas. $$ x^{2}+y^{2}+8 x-2 y-8=0 $$

Complete each solution to solve the system. Solve: \(\left\\{\begin{array}{l}x^{2}+y^{2}=5 \\ y=2 x\end{array}\right.\) $$ \begin{aligned} x^{2}+y^{2} &=5 \\ x^{2}+(\square)^{2} &=5 \\ x^{2}+4 x^{2} &=\square\\\ \square x^{2} &=5 \\ x^{2} &=\square \end{aligned} $$ $$ x=1 \quad \text { or } \quad x=-1 $$ If \(x=1,\) then \(y=2(\square)=2\) If \(x=-1,\) then \(y=2(\square)=-2\) The solutions are \((1,2)\) and \((-1, \square)\)

Solve each system of equations by elimination for real values of \(x\) and \(y .\) See Example 4 $$ \left\\{\begin{array}{l} 2 x^{2}+y^{2}=6 \\ x^{2}-y^{2}=3 \end{array}\right. $$

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} y=x+1 \\ x^{2}-y^{2}=1 \end{array}\right. $$

Show that the equations of the extended diagonals of the fundamental rectangle of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) are $$y=\frac{b}{a} x \quad \text { and } \quad y=-\frac{b}{a} x$$
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