Please help me find an equation of x and y that has the same graph as the polar equation r - 4 cos theta = 0. I need to show all of my work.

Question

Please help me find an equation using x and y that has the same graph as the polar r - 4 cos theta = 0. I need to show all of my work. I have tried numerous things; and nothing seems to work.

 

Answer 1

r - 4cosθ = 0

 

          r = 4cosθ

 

Multiply both sides by r:  r² = 4rcosθ

 

rcosθ = x and r² = x²+y²

 

So, x²+y² = 4x

 

     (x²-4x) + y² = 0

 

     Complete the square to get (x²-4x+4) + y² = 4

 

                                                (x-2)² + (y-0)² = 2²

 

      This is a circle centered at (2,0) with radius 2.  

 

Answer 2

Hello Ann.

 

When converting polar to Cartesian coordinates, remember that

 

r = √(x²+y²)

r*cos(θ) = x, and

r*sin(θ) = y.

 

For the above: substitute  √(x²+y²) for r and x/r = x/√(x²+y²) for cos(θ) to get:

√(x²+y²) - 4x/√(x²+y²) = 0. Multiply by √(x²+y²) to get:

x²+y²-4x = 0. Now you can solve for y:

y = √(4x-x²) or y = -√(4x-x²), where 0≤x≤4 (since what's inside the radical must be greater than or equal to 0).

 

You can also "complete the square" for x²+y²-4x = 0: 

x²-4x+4+y² = 4 (added 4 to both sides of the equation)

x²-4x+4+y² = (x-2)²+y² = 4 = 2².

Do you recognize the above as the equation for a circle of radius 2 centered at (2,0)?