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Determine whether each statement is true or false. Describe the graph (if it exists) of $$x^{2}+y^{2}+10 x-6 y+34=0$$

Short Answer

Expert verified
This is a point at (-5, 3); the statement is true.

Step by step solution

01

Rewriting the Equation

We start with the given equation: \(x^{2} + y^{2} + 10x - 6y + 34 = 0\). Our goal is to write it in a form that is easy to analyze, typically the equation of a circle or other conic.
02

Completing the Square for x

For the \(x\) terms, we have \(x^2 + 10x\). To complete the square, we take half of 10, which is 5, and square it to get 25. We add and subtract 25 in the equation:\[(x^2 + 10x + 25) - 25\].This can be rewritten as \((x + 5)^2 - 25\).
03

Completing the Square for y

Similarly, for the \(y\) terms, we have \(y^2 - 6y\). Taking half of -6 gives -3, and squaring it gives 9. Add and subtract 9:\[(y^2 - 6y + 9) - 9\].This rewrites as \((y - 3)^2 - 9\).
04

Substituting Back and Simplifying

Substitute the completed squares back into the original equation:\((x + 5)^2 - 25 + (y - 3)^2 - 9 + 34 = 0\).Simplify this to \((x + 5)^2 + (y - 3)^2 = 0\).
05

Analyzing the Result

The equation \((x + 5)^2 + (y - 3)^2 = 0\) implies that both squared terms must be zero, so \(x + 5 = 0\) and \(y - 3 = 0\). Solving these gives \(x = -5\) and \(y = 3\). Thus, this is a point at \((-5, 3)\) rather than a true circle, as the radius \(r = 0\).

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Key Concepts

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

Equation of a Circle
An equation of a circle is a fundamental concept in geometry and algebra. A standard form of a circle's equation is \[ (x - h)^2 + (y - k)^2 = r^2 \]where:
  • \( h \) and \( k \) are the coordinates of the center of the circle,
  • \( r \) is the radius.
The equation essentially accounts for all points \((x, y)\) at a fixed distance, \( r \), from the center \((h, k)\).
Completing the square is a typical method used to transform complex quadratic equations into a circle's standard form. This involves reorganizing and manipulating the equation so that both \(x\) and \(y\) terms form perfect squares.
In the given exercise, the simplification of the initial equation, through completing the square for both \(x\) and \(y\), showed that the result was not a circle but a point at \((-5, 3)\). This happens when the squared terms equal zero, indicating that the radius \(r\) is zero.
Conic Sections
Conic sections represent different shapes that arise from slicing a cone with a plane. They consist of circles, ellipses, parabolas, and hyperbolas—each with its own unique equation.
A circle is a special type of ellipse where the distance from the center to any point on the perimeter is constant (the radius).
The canonical forms help in identifying each conic section:
  • Circle: \( (x - h)^2 + (y - k)^2 = r^2 \)
  • Ellipse: \( \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \)
  • Parabola: \( y - k = a(x - h)^2 \) (or its horizontal counterpart)
  • Hyperbola: \( \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \)
When analyzing equations, recognizing conic sections aids in visualizing and understanding their geometric shapes.
In this exercise, the initial equation appeared quadratic, leading us to examine if it could transform into any of these sections. Ultimately, it simplified to a degenerate circle, which is actually a single point.
Analyzing Equations
Analyzing equations involves breaking down given expressions to determine their geometric and algebraic properties. This is essential in graphing them and understanding the forms they represent.
When presented with equations, especially quadratic ones like the given \(x^2 + y^2 + 10x - 6y + 34 = 0\), seeking to complete the square proves valuable. This method helps group terms to form recognizable patterns, such as perfect square trinomials.
During such analysis:
  • Identify if the equation represents a conic section.
  • Use algebraic manipulation to rewrite the equation in standard forms (e.g., completing the square).
  • Solve the resulting equation to find specific points or dimensions like the center and radius of a circle.
In the problem here, the analysis unveiled not a circle of finite radius, but a degenerate form, simplifying to a point at a specific location, \((-5, 3)\), due to a radius of zero.

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Most popular questions from this chapter

Explain the mistake that is made. Find the slope of the line that passes through the points (-2,3) and (4,1) Solution: Write the slope formula. \(\quad m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) Substitute (-2,3) and \((4,1), \quad m=\frac{1-3}{-2-4}=\frac{-2}{-6}=\frac{1}{3}\) This is incorrect. What mistake was made?

a. Solve the equation \(x^{2}-2 x=b, b=8\) by first writing in standard form. Now plot both sides of the equation in the same viewing screen \(\left(y_{1}=x^{2}-2 x \text { and } y_{2}=b \right)\). At what. \(x\) -values do these two graphs intersect? Do those points agree with the solution set you found? b. Repeat (a) for \(b=-3,-1,0,\) and 5

Find an equation of a line that passes through the point \((-A, B-1)\) and is perpendicular to the line \(A x+B y=C\) Assume that \(A\) and \(B\) are both nonzero.

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Refer to the following: Einstein's special theory of relativity states that time is relative: Time speeds up or slows down, depending on how fast one object is moving with respect to another. For example, a space probe traveling at a velocity \(v\) near the speed of light \(c\) will have "clocked" a time \(t\) hours, but for a stationary observer on Earth that corresponds to a time \(t_{0} .\) The formula governing this relativity is given by $$ t=t_{0} \sqrt{1-\frac{v^{2}}{c^{2}}} $$ If the time elapsed on a space probe mission is 5 years but the time elapsed on Earth during that mission is 30 years, how fast is the space probe traveling? Give your answer relative to the speed of light.

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