140k views
4 votes
Determine center point and radius in the equation x^2+y^2-6y+1=0

User Dsnunez
by
8.3k points

1 Answer

3 votes

9514 1404 393

Answer:

  • center: (0, 3)
  • radius: 2√2

Explanation:

The equation can be put into standard form by "completing the square" for each variable. The x-term is already a square, so we need do this only for the y-term.

x^2 +(y^2 -6y) = -1

x^2 +(y^2 -6y +9) = -1 +9 . . . . . . add (-6/2)^2 to complete the square

x^2 +(y -3)^2 = 8 . . . . standard form equation

__

Comparing the above equation to the form:

(x -h)^2 +(y -k)^2 = r^2 . . . . . . . circle centered at (h, k), radius r

we see that (h, k) = (0, 3) and r^2 = 8 ⇒ r = 2√2.

The center point is (0, 3); the radius is 2√2.

User Tubbe
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories