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

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The polar equation r = 7cos(θ) can be converted into a rectangular equation by using the relationships between polar and rectangular coordinates. Starting with r = 6(cos(θ)), we substitute cos(θ) with x/r, leading to the equation r^2 = 6x. By substituting r^2 with x^2 + y^2, we arrive at x^2 + y^2 = 6x. Rearranging this into standard form, we complete the square to find that the equation represents a circle with a radius of 3 units, centered at the point (3, 0). This conversion illustrates the relationship between polar coordinates and their rectangular counterparts effectively.
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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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