97.3k views
3 votes
What is the center of a circle whose equation is x^2+y^2-12x-2y+12=0

What is the center of a circle whose equation is x^2+y^2-12x-2y+12=0-example-1

1 Answer

6 votes

Answer:

The center of this circle is at (6, 1).

Explanation:

Rewrite x^2+y^2-12x-2y+12=0 by grouping like terms. Then:

x^2+y^2-12x-2y+12=0 becomes x^2 - 12x +y^2 - 2y + 12=0.

Next, complete the squares:

x^2 - 12x + 36 - 36 + y^2 - 2y + 1 - 1 + 12 = 0.

Rewriting the two perfect squares as squares of binomials, we get:

(x - 6)^2 - 36 + (y - 1)^2 - 1 + 12 = 0

Moving the constants to the right side:

(x - 6)^2 + (y - 1)^2 = 36 + 1 - 12 = 25

Then the desired equation is:

(x - 6)^2 + (y - 1)^2 = 5^2. The center of this circle is at (6, 1).

User Dinesh Kaushik
by
8.1k 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