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Chapter 8: Problem 48

Decide whether each equation has a circle as its graph. If it does, give the center and radius. $$x^{2}-12 x+y^{2}+10 y=-25$$

Short Answer

Expert verified
The equation represents a circle with center (6, -5) and radius 6.

Step by step solution

01

Identify the General Equation of a Circle

A circle's general equation is \[ (x - h)^2 + (y - k)^2 = r^2 \]where \((h, k)\) is the center of the circle and \(r\) is the radius.
02

Rearrange the Given Equation

The given equation is \[ x^2 - 12x + y^2 + 10y = -25 \]Our goal is to rearrange it into the standard form of a circle by completing the square for both \(x\) and \(y\).
03

Complete the Square for x-Terms

For the \(x\)-terms, take the coefficient of \(x\), which is \(-12\), divide it by 2, and square it: \[ \left(\frac{-12}{2}\right)^2 = 36 \].Add and subtract 36 inside the equation:\[ x^2 - 12x + 36 - 36 \].
04

Complete the Square for y-Terms

For the \(y\)-terms, take the coefficient of \(y\), which is 10, divide it by 2, and square it: \[ \left(\frac{10}{2}\right)^2 = 25 \].Add and subtract 25 inside the equation:\[ y^2 + 10y + 25 - 25 \].
05

Form Perfect Squares

Rearrange terms to form perfect squares:\[ (x^2 - 12x + 36) + (y^2 + 10y + 25) - 36 - 25 = -25 \].Rewriting the perfect squares gives:\[ (x - 6)^2 + (y + 5)^2 = 36 \].
06

Identify the Center and Radius

Now that the equation is in the form \[ (x - h)^2 + (y - k)^2 = r^2 \],we identify the center \((h, k) = (6, -5)\) and the radius \(r = 6\), because \[ 36 = r^2 \].

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Key Concepts

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

Completing the Square
Completing the square is a technique used to convert a quadratic equation into a perfect square trinomial, which makes it easier to analyze. In the context of circle equations, it helps us transform the quadratic expressions into a recognizable circle form. To complete the square for a term like \(x^2 - 12x\):
  • Identify the coefficient of \(x\), which is -12 in this case.
  • Divide by 2 to get -6, and then square this result to get 36.
  • Add and subtract this new number (36) within the equation to form a complete square.
For \(y^2 + 10y\), repeat the same process:
  • Take the coefficient of \(y\), which is 10.
  • Divide by 2 to get 5, and square to get 25.
  • Add and subtract 25 in the equation.
This process changes the quadratic terms into perfect square trinomials, such as \((x-6)^2\) and \((y+5)^2\). This format reveals the geometric properties of the circle more clearly.
Center of a Circle
The center of a circle in a standard circle equation \((x - h)^2 + (y - k)^2 = r^2\) is given by the point \((h, k)\). Completing the square helps us put the equation into this form, making it easy to identify the center. Looking at the terms:
  • \((x-6)^2\) indicates a horizontal shift to the right by 6 units.
  • \((y+5)^2\) shows a vertical shift downwards by 5 units, as it can be defined as \((y - (-5))^2\).
Thus, the center of the circle described in this equation is at \((6, -5)\). This center point is crucial as it defines the position of the circle in the coordinate plane.
Radius of a Circle
The radius \(r\) of a circle can be directly identified from the standard equation of a circle \((x - h)^2 + (y - k)^2 = r^2\). In our problem:
  • The equation was transformed into \((x - 6)^2 + (y + 5)^2 = 36\).
  • The right side of the equation is \(r^2\), thus \(36 = r^2\).
  • This means the radius \(r = \sqrt{36} = 6\).
The radius represents the distance from the center of the circle to any point on its circumference. In this example, the circle has a radius of 6 units. Knowing the radius allows us to understand the size of the circle, which helps in graphing and visualizing its position in geometric problems.

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Most popular questions from this chapter

Find an equation for each hyperbola. \(x\) -intercepts \((\pm 3,0) ;\) foci \((\pm 4,0)\)

Equation of an Ellipse Use the definition of an ellipse to find an equation of an ellipse with foci \((3,0)\) and \((-3,0),\) where the sum of the distances from any point of the ellipse to the two foci is \(10 .\)

Find a rectangular equation. State the appropriate interval for \(x\) or \(y .\) $$x=\sqrt{t}, y=t^{2}-1, \text { for } t \text { in }[0, \infty)$$

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