Find the center and the radius for each of the circles x^2-2x+y^2-4y-4=0 and x^2-2x+y^2-4y+5=0.

Question

How many points fit the equation x^2-2x+y^2-4y+9=0?

Answer 1

PART 1:  x^2 -2x + y^2 -4y - 4 = 0

 

  Move the constant to the right side of the equals sign.

 

     x^2 - 2x + y^2 -4y = 4

 

  Use completing the square on the x terms.  b = -2 ... take half of that, square it, and add the result to both sides.

 

    x^2 - 2x + 1 + y^2 - 4y = 4 + 1

 

   Simplify the right side, and factor the x term trinomial

 

    (x - 1)^2 + y^2 - 4y = 5

 

   Use completing the square on the y terms.  b = -4 ... take half of that, square it, and add the result to both sides.

 

    (x - 1)^2 + y^2 -4y + 4 = 5 + 4

 

   Simplify the right side, and factor the y term trinomial

 

    (x - 1)^2 + (y - 2)^2 = 3^2

 

   Thus this is a circle centered at (1,2) with a radius of 3

 

PART 2:  x^2 - 2x + y^2 - 4y + 5 = 0

 

  Move the constant to the right side of the equals sign

 

     x^2 - 2x + y^2 - 4y = -5

 

  Use completing the square on the x terms.  b = -2 ... take half of that, square it, and add the result to both sides.

 

     x^2 - 2x + 1 + y^2 - 4y = -5 + 1

 

  Simplify the right side of the equals, and factor the x term trinomial

 

    (x - 1)(x - 1) + y^2 - 4y = -4

 

      Use complete the square on the y terms.  b = -4 ... take half of that, square it, and add the result to both sides.

 

    (x-1)^2 + y^2 -4y + 4 = -4 + 4

 

   Simplify the right side of the equals, and factor the y term trinomial

 

    (x-1)^2 + (y - 2)^2 = 0

 

   Since the radius is 0 ... this results in a graph that shows point (1,2)