/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 82 Find the center-radius form of t... [FREE SOLUTION] | 91影视

91影视

Find the center-radius form of the circle with the given equation. Determine the coordinates of the center, find the radius, and graph the circle. $$x^{2}+y^{2}-10 x+6 y+21=0$$

Short Answer

Expert verified
Center: (5, -3); Radius: \(\sqrt{13}\); in form \((x-5)^2 + (y+3)^2 = 13\).

Step by step solution

01

Move the Constant Term

First, we rewrite the equation by moving the constant term to the other side of the equation:\[ x^2 + y^2 - 10x + 6y = -21 \]
02

Group and Prepare to Complete the Square

Group the \(x\) terms together and the \(y\) terms together: \[ (x^2 - 10x) + (y^2 + 6y) = -21 \] Now, we will complete the square for both groups.
03

Complete the Square for x

To complete the square for the \(x\) terms, take half of the coefficient of \(x\) (which is \(-10\)), square it, and add it inside the bracket. The half is \(-5\), and its square is \(25\). Add \(25\) inside the brackets for \(x\): \[ (x^2 - 10x + 25) + (y^2 + 6y) = -21 + 25 \] This makes the \(x\) group a perfect square: \[ (x-5)^2 \]
04

Complete the Square for y

For the \(y\) terms, take half of the coefficient of \(y\) (which is \(6\)), square it, and add it inside the bracket. The half is \(3\), and its square is \(9\). Add \(9\) inside the brackets for \(y\): \[ (x-5)^2 + (y^2 + 6y + 9) = -21 + 25 + 9 \] This makes the \(y\) group a perfect square: \[ (x-5)^2 + (y+3)^2 = 13 \]
05

Write the Center-Radius Form

Now, the equation is in the form of \[ (x-h)^2 + (y-k)^2 = r^2 \]where \((h, k)\) is the center and \(r\) is the radius. Therefore, for this equation:- Center \((h, k)\): \((5, -3)\)- Radius \(r\): \(\sqrt{13}\)
06

Graph the Circle

To graph the circle, plot the center at \((5, -3)\). Then, with a compass set to the length of the radius \(\sqrt{13} \approx 3.6\), draw a circle around the center point. This visual should encapsulate the entire equation on the coordinate plane.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Center-Radius Form
The center-radius form of a circle's equation is a neat way to express the circle's most important features: its center and its radius. This form is written as \((x-h)^2 + (y-k)^2 = r^2\), where
  • \((h, k)\) represents the center of the circle.
  • \(r\) is the radius of the circle.
This format is extremely useful because it displays at a glance the position of the circle on the coordinate plane and how big the circle is.
For the equation in the given exercise,
  • The center is at \((5, -3)\).
  • The radius \(r\) is \(\sqrt{13}\).
The center-radius formis an excellent tool for graphing and understanding the spatial relationship of circles in geometry.
Completing the Square
Completing the square is a critical technique in coordinate geometry that helps us find the center and radius of a circle from its equation. When you encounter an equation like \(x^2 + y^2 - 10x + 6y + 21 = 0\),we can simplify and transform it using this method. The key steps to complete the square are:
  • Group the \(x\) terms and \(y\) terms separately.
  • Find the necessary number to add inside each group to make it a perfect square trinomial.鈥
  • Apply the formula \((x-a)^2 = x^2 - 2ax + a^2\).
For example, the \(x\) terms, \(x^2 - 10x\), need \(( -5 )^2 = 25\).Similarly, the \(y\) terms,\(y^2 + 6y\), require \((3)^2 = 9\).
By adding these to the equation, both \((x-5)^2\) and \((y+3)^2\) form clean squared terms, ultimately simplifying it to the center-radius form.
Coordinate Geometry
Coordinate geometry provides a way to analyze geometric figures like circles using an algebraic approach, which involves the use of the Cartesian coordinate system. A circle in coordinate geometry is typically represented by its equation in the center-radius form. This allows us to identify crucial elements such as:
  • The center of the circle, which determines the circle's location in the plane.
  • The radius, which tells us how big the circle is.
By using coordinate geometry, you gain insights into how circles interact with lines and other shapes.
It also facilitates transformations and graphing, giving you powerful ways to visualize and solve problems involving circles.
Graphing Circles
Graphing a circle begins with identifying the circle's center and radius from its equation. Once you have the equation in the center-radius form, \((x-h)^2 + (y-k)^2 = r^2\),you can easily pinpoint these features. Here's how to graph a circle in simple steps:
  • Locate the center at \((h, k)\) on the coordinate plane.
  • Measure the radius \(r\) from the center using a ruler or compass鈥擻(r\) equals the square root of the value on the right side of the equation.
  • Draw a circle around the center with that radius distance, ensuring that the curve is smooth and even.
In our case, the center is at \((5, -3)\), and the radius is \(\sqrt{13} \approx 3.6\). By following these steps, you create a precise visual representation that can be used to solve further geometric problems or understand more complex concepts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve each inequality analytically. Support your answers graphically. Give exact values for endpoints. (a) \(x^{2}+4 x+3 \geq 0\) (b) \(x^{2}+4 x+3<0\)

Solve each equation. For equations with real solutions, support your answers graphically. $$\frac{2}{3} x^{2}+\frac{1}{4} x=3$$

For each pair of numbers, find the values of \(a, b,\) and \(c\) for which the quadratic equation ax \(^{2}+b x+c=0\) has the given numbers as solutions. Answers may vary. (Hint: Use the zero-product property in reverses $$2-\sqrt{5}, 2+\sqrt{5}$$

Solve each problem. The table lists the average heating bill for a natural gas consumer in Indiana during various months of the year. $$\begin{array}{|c|c|} \hline \text { Month } & \text { Bill ( } \$ \text { ) } \\ \hline \text { Jan. } & 108 \\ \text { Mar. } & 68 \\ \text { May } & 18 \\ \text { July } & 12 \\ \text { Sept. } & 13 \\ \text { Nov. } & 54 \\\\\hline \end{array}$$ (a) Plot the data. Let \(x=1\) correspond to January, \(x=2\) to February, and so on. (b) Find a quadratic function \(f(x)=a(x-h)^{2}+k\) that models the data. Use \((7,12)\) as the vertex and \((1,108)\) as another point to determine \(a\) (c) Plot the data together with the graph of \(f\) in the same window. How well does \(f\) model the average heating bill over these months? (d) Use the quadratic regression feature of a graphing calculator to determine the quadratic function \(g\) that provides the best fit for the data. (e) Use the functions \(f\) and \(g\) to approximate the heating bill to the nearest dollar in the following months. (i) February (ii) June

For each pair of numbers, find the values of \(a, b,\) and \(c\) for which the quadratic equation ax \(^{2}+b x+c=0\) has the given numbers as solutions. Answers may vary. (Hint: Use the zero-product property in reverses $$2 i,-2 i$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.