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.
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).
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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