MHB Convert r=7cos(theta) into a rectangular equation

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SUMMARY

The polar equation r=7cos(θ) can be converted into a rectangular equation using the relationships between polar and rectangular coordinates. By substituting r and θ with x and y, the equation transforms into x² + y² = 6x. Further manipulation leads to the standard form (x-3)² + y² = 3², indicating a circle with a radius of 3 units centered at the point (3,0). This conversion process is essential for understanding the geometric representation of polar equations in Cartesian coordinates.

PREREQUISITES
  • Understanding of polar coordinates and their relationship to rectangular coordinates
  • Familiarity with the equations r = x/cos(θ) and r = y/sin(θ)
  • Knowledge of completing the square in algebra
  • Basic concepts of conic sections, specifically circles
NEXT STEPS
  • Study the conversion of other polar equations to rectangular form
  • Learn about the properties of circles in standard form
  • Explore the graphical representation of polar equations
  • Investigate the applications of polar coordinates in calculus
USEFUL FOR

Students learning algebra and geometry, educators teaching coordinate systems, and anyone interested in the graphical representation of mathematical equations.

Elissa89
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So we're learning to plot polar equations, which easy enough. But I got a question in the homework that wasn't covered in class:

Convert r=7cos(theta) into a rectangular equation. Use x and y values. I know how to convert when it's x=r*cos(theta) or y=r*sin(theta) and r and theta is given. But this is different and I don't know how to do it.
 
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Okay, we are given the polar equation:

$$r=6\cos(\theta)$$

Now, from:

$$x=r\cos(\theta)\implies \cos(\theta)=\frac{x}{r}$$

We may write:

$$r=6\left(\frac{x}{r}\right)$$

Multiply through by \(r\):

$$r^2=6x$$

We know:

$$r^2=x^2+y^2$$

Hence, we have:

$$x^2+y^2=6x$$

This would technically suffice, but I would prefer to continue and put into standard form:

$$x^2-6x+y^2=0$$

Complete the square on \(x\):

$$(x-6x+9)+y^2=9$$

$$(x-3)^2+y^2=3^2$$

Now it's easy to see we have a circle of radius 3 units centered at (3,0).
 

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