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Chapter 13: Problem 29

Solve. Remember that graphs can be used to confirm all real solutions. $$ \begin{aligned} &x^{2}+y^{2}=9\\\ &x^{2}-y^{2}=9 \end{aligned} $$

Short Answer

Expert verified
The solutions are (3, 0) and (-3, 0).

Step by step solution

01

Understand the given equations

The exercise gives two equations: 1) The first equation is a circle equation: \( x^{2} + y^{2} = 9 \) 2) The second equation is a hyperbola equation: \( x^{2} - y^{2} = 9 \)
02

Add the two equations

To simplify the process, add the two given equations: \( x^{2} + y^{2} = 9 \) \( x^{2} - y^{2} = 9 \) Adding these equations results in: \( (x^{2} + y^{2}) + (x^{2} - y^{2}) = 9 + 9 \) \( 2x^{2} = 18 \) Simplify: \( x^{2} = 9 \) Take the square root of both sides: \( x = \text{卤} 3 \)
03

Solve for y with each x value

Substitute \( x = 3 \) into the first equation: \( 3^{2} + y^{2} = 9 \) \( 9 + y^{2} = 9 \) \( y^{2} = 0 \) Thus, \( y = 0 \). Similarly, substitute \( x = -3 \) into the first equation: \( (-3)^{2} + y^{2} = 9 \) \( 9 + y^{2} = 9 \) \( y^{2} = 0 \) Thus, \( y = 0 \).
04

Confirm Solutions

The solutions are the points \( (3, 0) \) and \( (-3, 0) \).
05

Graphical Confirmation

Graph both equations on the same coordinate plane to see the points of intersection. The circle \( x^{2} + y^{2} = 9 \) and the hyperbola \( x^{2} - y^{2} = 9 \) intersect at the points \( (3, 0) \) and \( (-3, 0) \).

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Key Concepts

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

Circle Equation
In mathematics, the equation of a circle is fundamental when dealing with geometric shapes. The standard form of a circle's equation is given by \( (x - h)^{2} + (y - k)^{2} = r^{2} \), where \( (h, k) \) represents the coordinates of the circle's center, and \( r \) is the radius. In our given problem, the circle's equation is \( x^{2} + y^{2} = 9 \). This means the circle is centered at the origin (0,0) with a radius of \sqrt{9} = 3\.
  • The term \( x^{2} \) and \( y^{2} \) shows the distances from the center to any point on the circle remains constant.

  • Adding or subtracting the same values to these squared terms shifts the circle around the coordinate plane accordingly.
Knowing this helps us visualize and understand the graph of the circle that passes through (3,0), (-3,0), (0,3), and (0,-3).
Hyperbola Equation
Just as the circle has its standard equation, so does a hyperbola. The standard form of a hyperbola's equation is usually given by \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \) for horizontal hyperbolas or \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \) for vertical hyperbolas.
In our problem, we have \( x^{2} - y^{2} = 9 \), which aligns with the horizontal hyperbola centered at the origin (0,0) and transverse axis along the x-axis. To match the given equation with the standard form, we can rewrite it as \( \frac{x^{2}}{9} - \frac{y^{2}}{9} = 1 \).
  • This tells us that \ a^{2} = 9 \ and \right and left-transversal axis length is 2a = 6,
  • and b=b=3 indicating vertical with asymptotes y=卤3
    • This information is key for drawing the hyperbola and understanding its intersections with other shapes.
Graphical Confirmation
Graphical confirmation is a powerful step in solving and verifying solutions to equations. By plotting both the circle \( x^{2} + y^{2} = 9 \) and the hyperbola \( x^{2} - y^{2} = 9 \) on the same coordinate plane, we can visually confirm the points of intersection.
Here鈥檚 how we do it:
  • First, we draw the circle with center (0,0) and radius 3.
  • Next, we draw the hyperbola, also centered at (0,0) with a transverse axis of 6 and asymptotes along the lines \ y = \ 鉁�3.
The intersections will be at these 2 points (3,0) and (-3,0).
This graphical method gives a clear visual representation, confirming the accuracy of algebraic solutions.

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Most popular questions from this chapter

Find an equation of the circle satisfying the given conditions. Center \((-5,-8),\) radius \(3 \sqrt{2}\)

Solve. \(3 x^{2}+10 x-8=0\)

Match the equation with the center or vertex of its graph, listed in the column on the right. a) Vertex: \((-2,5)\) b) Vertex: \((5,-2)\) c) Vertex: \((2,-5)\) d) Vertex: \((-5,2)\) e) Center: \((-2,5)\) f) Center: \((2,-5)\) g) Center: \((5,-2)\) h) Center: \((-5,2)\) $$x=(y-5)^{2}-2$$

A rectangle has an area of \(2 \mathrm{yd}^{2}\) and a perimeter of 6 yd. Find its dimensions.

Find the center and the radius of each circle. Then graph the circle. $$ x^{2}+y^{2}+7 x-3 y-10=0 $$
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