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Chapter 1: Problem 17

In Exercises 15 through 20 , determine whether the graph is a circle, a point- circle, or the empty set. $$ x^{2}+y^{2}-10 x+6 y+36=0 $$

Short Answer

Expert verified
The graph is the empty set.

Step by step solution

01

- Identify general form

Recognize that the given equation is in the general form: \[ x^2 + y^2 + Dx + Ey + F = 0 \]For this problem: \[ x^2 + y^2 - 10x + 6y + 36 = 0 \]Thus, \(D = -10\), \(E = 6\), and \(F = 36\).
02

- 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 and subtract this value inside the equation. \(\left( \frac{-10}{2} \right)^2 = (-5)^2 = 25\)Adding and subtracting 25 gives: \[ x^2 - 10x + 25 - 25 \]Rewrite this as: \[ (x - 5)^2 - 25 \]
03

- Complete the square for y

To complete the square for the \(y\) terms, take half of the coefficient of \(y\) (which is 6), square it, and add and subtract this value inside the equation. \(\left( \frac{6}{2} \right)^2 = 3^2 = 9\)Adding and subtracting 9 gives: \[ y^2 + 6y + 9 - 9 \]Rewrite this as: \[ (y + 3)^2 - 9 \]
04

- Rewrite the equation

Substitute the completed squares for \(x\) and \(y\) back into the equation: \[ (x - 5)^2 - 25 + (y + 3)^2 - 9 + 36 = 0 \]
05

- Simplify the equation

Combine constants on the right side of the equation: \[ (x - 5)^2 + (y + 3)^2 - 25 - 9 + 36 = 0 \]\[ (x - 5)^2 + (y + 3)^2 + 2 = 0 \]Subtract 2 from both sides: \[ (x - 5)^2 + (y + 3)^2 = -2 \]
06

- Determine the nature of the graph

Since the right side of the equation \[ (x - 5)^2 + (y + 3)^2 = -2 \] is negative, it indicates that there is no real solution. Therefore, the graph represents the empty set.

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Key Concepts

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

Circle Equations
In coordinate geometry, a circle's equation in the standard form is written as \[(x - h)^2 + (y - k)^2 = r^2\], where *(h, k)* is the center of the circle and \(r\) is the radius. This equation is obtained by completing the square for the terms involving x and y.For example, consider an equation like \[x^2 + y^2 - 10x + 6y + 36 = 0\]. To transform this into standard form, first complete the square for the x terms and y terms separately. The completed squares reveal the center and radius. However, if you end up with a negative value on the right side of the equation, like -2 in this problem, it means the graph represents an empty set, because it's impossible to have a negative radius squared.
Graph Analysis
Graph analysis involves understanding the nature of the graph from its equation. For circle equations, a crucial aspect is determining whether the equation represents a real circle, a point-circle, or the empty set.
  • A real circle occurs if the equation matches the form \[(x - h)^2 + (y - k)^2 = r^2\] with a positive r.
  • A point-circle represents a degenerate case where the radius r is zero, reducing the equation to \[(x - h)^2 + (y - k)^2 = 0\]. This signifies a single point (h, k).
  • The empty set happens if the right-hand side is negative, as in \[(x - 5)^2 + (y + 3)^2 = -2\]. No real values of x and y will satisfy this equation, so no graph exists.
Coordinate Geometry
Coordinate geometry is the study of geometry using a coordinate system. Here, we use the Cartesian coordinate system where each point on the plane is identified by an ordered pair (x, y).
  • To derive the circle's equation from its general form \[Ax^2 + Ay^2 + Dx + Ey + F = 0\], complete the square for both x and y terms.
  • The process of completing the square helps convert an equation into a recognizable geometric form.
  • The coordinates of the center and the radius of a circle can be identified from the completed square form: \[(x - h)^2 + (y - k)^2 = r^2\].
By understanding these core principles, you can tackle a wide range of problems in coordinate geometry, providing a solid foundation for further studies.

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Most popular questions from this chapter

In Exercises 1 through 10 , find the domain and range of the given function, and draw a sketch of the graph of the function. $$ H=\\{(x, y)|y=| x-3 \mid\\} $$

In Exercises 1 through 10, list the elements of the given set if \(A=\\{0,2,4,6,8\\}, B=\\{1,2,4,8\\}, C=\\{1,3,5,7,9\\}\), and \(D=\) \(\\{0,3,6,9\\}\). $$ B \cap D $$

In Exercises 1 through 10, solve for \(x\). $$ |5-2 x|=11 $$

In Exercises 11 through 14 , find the center and radius of each circle, and draw a sketch of the graph. $$ x^{2}+y^{2}-10 x-10 y+25=0 $$

Prove analytically that a line from the center of any circle bisecting any chord is perpendicular to the chord.
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