Write the equation of the circle x^2+y^2+6x-2y-6=0 in the standard form.

2 Answers | Add Yours

aliddi's profile pic

aliddi | High School Teacher | (Level 1) Adjunct Educator

Posted on

Given: x^2 + y ^2 + 6x - 2y - 6 = 0

1. Group the x's together, group the y's together: x^2 + 6x + y^2 - 2y - 6 = 0.

2. Move the numbers to the other side of the equation (add 6): x^2 + 6x + y^2 - 2y = 6.

3. Complete the squares for the 'x' and 'y' terms (halve the coefficient of the x/y term, then square it. Because we are adding this to the left hand side of the equation, we must also add it to the right-hand side in order to maintain equality): x^2 + 6x + 9 + y^2 - 2y + 1= 6 + 9 + 1. --> x^2 + 6x + 9 + y^2 - 2y + 1 = 16.

4. Factor the perfect squares you created in the above step: (x + 3)^2 + (y - 1)^2 = 16.

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to write the equation of the circle x^2+y^2+6x-2y-6=0 in the standard form.

x^2+y^2+6x-2y-6=0

=> x^2 + 6x + y^2 - 2y - 6 = 0

=> x^2 + 6x + 9 + y^2 - 2y + 1 - 6 - 1 - 9 = 0

=> (x + 3)^2 + (y - 1)^2 - 6 - 1 - 9 = 0

=> (x + 3)^2 + (y - 1)^2 = 16

=> (x + 3)^2 + (y - 1)^2 = 4^2

The standard form of the circle is (x + 3)^2 + (y - 1)^2 = 4^2

We’ve answered 318,947 questions. We can answer yours, too.

Ask a question