Chapter 10: Problem 22
In Problems 17-22, find the Cartesian equations of the graphs of the given polar equations. \(r^{2}-6 r \cos \theta-4 r \sin \theta+9=0\)
Short Answer
Expert verified
The Cartesian equation is \((x - 3)^2 + (y - 2)^2 = 4\).
Step by step solution
01
Understand the Problem
We need to convert the polar equation \(r^2 - 6r \cos \theta - 4r \sin \theta + 9 = 0\) into a Cartesian equation. This involves using the relationships between polar and Cartesian coordinates: \(x = r \cos \theta\), \(y = r \sin \theta\), and \(r = \sqrt{x^2 + y^2}\).
02
Substitute Polar to Cartesian
Replace \(r \cos \theta\) with \(x\) and \(r \sin \theta\) with \(y\). The equation becomes \(r^2 - 6x - 4y + 9 = 0\).
03
Express \(r^2\) in Terms of \(x\) and \(y\)
Use the identity \(r^2 = x^2 + y^2\). Substitute \(r^2\) with \(x^2 + y^2\) in the equation yielding \(x^2 + y^2 - 6x - 4y + 9 = 0\).
04
Complete the Square
Complete the square for both \(x\) and \(y\):1. For \(x\): \(x^2 - 6x = (x - 3)^2 - 9\).2. For \(y\): \(y^2 - 4y = (y - 2)^2 - 4\).
05
Combine and Simplify
Substitute the completed squares back into the equation:\((x - 3)^2 - 9 + (y - 2)^2 - 4 + 9 = 0\). Simplifying gives \((x - 3)^2 + (y - 2)^2 = 4\).
06
Identify the Result
The Cartesian equation \((x - 3)^2 + (y - 2)^2 = 4\) is a standard circle equation centered at \((3, 2)\) with a radius of 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Coordinates
Polar coordinates are often used in mathematics to describe locations on a plane using angles and radii. Unlike Cartesian coordinates, which use x and y positions, polar coordinates express a point by considering its distance from a reference point (the origin) and the angle relative to a reference direction, usually the positive x-axis. This system is particularly useful for problems involving circles and arcs.
To convert a point from Cartesian to polar coordinates, we use the formulas:
To convert a point from Cartesian to polar coordinates, we use the formulas:
- The radius, \(r\), is calculated as \(r = \sqrt{x^2 + y^2}\)
- The angle, \(\theta\), is found using \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\)
Converting Equations
Converting an equation from polar to Cartesian coordinates involves substituting the polar terms with their Cartesian equivalents. In our problem, the polar equation given is \(r^2 - 6r \cos \theta - 4r \sin \theta + 9 = 0\).
We know that:
This methodical substitution transforms our polar equation into the Cartesian form we need.
We know that:
- \(r \cos \theta = x\)
- \(r \sin \theta = y\)
- \(r^2 = x^2 + y^2\)
This methodical substitution transforms our polar equation into the Cartesian form we need.
Completing the Square
Completing the square is a useful algebraic tool for transforming quadratic expressions into a perfect square trinomial, simplifying many mathematical processes, especially when working with circles and ellipses.
Given the transformation to the Cartesian equation \(x^2 + y^2 - 6x - 4y + 9 = 0\), we apply completing the square to both the \(x\) and \(y\) products.
Given the transformation to the Cartesian equation \(x^2 + y^2 - 6x - 4y + 9 = 0\), we apply completing the square to both the \(x\) and \(y\) products.
- For \(x\), we rewrite \(x^2 - 6x\) as \((x - 3)^2 - 9\).
- For \(y\), we rewrite \(y^2 - 4y\) as \((y - 2)^2 - 4\).
Circle Equation
The standard form of a circle's equation in Cartesian coordinates is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) represents the center of the circle, and \(r\) is the radius.
In our solution, by transforming the equation through completing the square, we achieved this form: \((x - 3)^2 + (y - 2)^2 = 4\).
From this equation, we can directly see that:
In our solution, by transforming the equation through completing the square, we achieved this form: \((x - 3)^2 + (y - 2)^2 = 4\).
From this equation, we can directly see that:
- The center of the circle is at the point \((3, 2)\).
- The radius is \(\sqrt{4} = 2\).