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Chapter 1: Problem 77

Find the center and radius of the circle, and sketch its graph. $$ x^{2}+y^{2}=25 $$

Short Answer

Expert verified
The center of the circle is (0, 0) and the radius is 5.

Step by step solution

01

Identify the circle equation

Recognize the given equation: \(x^{2}+y^{2}=25\). This is a circle equation in standard form \((x-h)^{2}+(y-k)^{2}=r^{2}\).
02

Find the center

The circle's center is given by the coordinates \(h\) and \(k\). In the given equation, you can notice there's no 'minus' or 'plus' operation on \(x\) and \(y\), so both \(h\) and \(k\) equal 0. So the center of the circle is at the origin, \((0, 0)\).
03

Find the radius

The radius can be found by taking the square root of the expressions on the right side of the equation, so, the radius \(r = \sqrt{25} = 5\).
04

Sketch the graph

Draw a circle on a graph at the center position \((0, 0)\) with radius 5. The circle's boundary should pass through the points (5,0), (0,5), (-5,0) and (0,-5).

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Key Concepts

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

Center of the Circle
In the equation of a circle, the center is a crucial element. For the circle equation \((x - h)^2 + (y - k)^2 = r^2\), the center is located at the point \((h, k)\). If the equation looks simpler, like \(x^2 + y^2 = 25\), it means there are no added or subtracted constants besides the variables. This corresponds to \(h = 0\) and \(k = 0\).
Therefore, the center is at the origin \((0, 0)\). It's like the heart of the circle, showing where the circle is perfectly balanced.
  • If you see no alterations in the \(x\) or \(y\) terms, the circle's center is at \( (0, 0) \).
  • Any shift from the origin would involve terms like \((x - 2)^2\) or \((y + 3)^2\) indicating the center at \((2, -3)\).
Radius of the Circle
The radius is another vital feature of a circle. It is the distance from the center to any point along the circle. In our standard equation \((x - h)^2 + (y - k)^2 = r^2\), \(r^2\) represents the square of the radius.
For the given equation \(x^2 + y^2 = 25\), the right side \(25\) is \(r^2\).Taking the square root gives the actual radius, so \(r = \sqrt{25} = 5\).
  • Think of the radius as the circle's reach. It's how far the circle stretches from the center.
  • This tells you how large the circle is and helps in sketching it accurately on a graph.
Standard Form of a Circle Equation
The standard form of a circle equation is neat and structured: \((x - h)^2 + (y - k)^2 = r^2\).This form helps identify the center and radius easily. Understanding this form clears up how circles are positioned in coordinate geometry.
Here's how it works:
  • The terms \((x - h)\) and \((y - k)\) indicate the shifts in \(x\) and \(y\) directions, identifying the center \((h, k)\).
  • The \(r^2\) represents how large the circle is.
  • Our example, \(x^2 + y^2 = 25\), shows a circle centered at the origin, simplifying to \((x - 0)^2 + (y - 0)^2 = 5^2\).
This form is friendly for conversion and analysis, allowing easy manipulation and understanding.

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Most popular questions from this chapter

(a) find the inverse function of \(f\), (b) graph both \(f\) and \(f^{-1}\) on the same set of coordinate axes, (c) describe the relationship between the graphs of \(f\) and \(f^{-1}\), and (d) state the domain and range of \(f\) and \(f^{-1}\). $$ f(x)=-\frac{2}{x} $$

Determine whether the function has an inverse function. If it does, find the inverse function. $$ f(x)=\left\\{\begin{array}{ll} -x, & x \leq 0 \\ x^{2}-3 x, & x>0 \end{array}\right. $$

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