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Find the center of a circle with the equation:
x² + y² + 2x -20y – 20 = 0

User Godvsdeity
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1 Answer

5 votes

Answer: Center = (-1, 10)

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Step-by-step explanation:

If we had x^2+2x + ___, then 1 must go in the blank so that we have x^2+2x+1 = (x+1)^2

So we must add 1 to both sides to complete the square for the x terms. To find this value '1' we take half of the x coefficient 2 to get 1, then square it to get 1^2 = 1.

We have

x^2+y^2 + 2x - 20y - 20 = 0

x^2+y^2 + 2x - 20y - 20 + 1 = 0 + 1

(x^2+2x+1) + y^2 - 20y - 20 = 1

(x^2+2x+1) + y^2 - 20y = 1 + 20

(x+1)^2 + y^2 - 20y = 21

After completing the square for the x terms. Repeat for the y terms. Take half of -20 to get -10, which squares to 100. Add this to both sides

(x+1)^2 + y^2 - 20y = 21

(x+1)^2 + y^2 - 20y + 100 = 21+100

(x+1)^2 + (y - 10)^2 = 121

The equation is in the form (x-h)^2 + (y-k)^2 = r^2 where

  • h = -1
  • k = 10
  • r = 11

The center is (h, k) = (-1, 10). The radius is r = 11.

User Brux
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