Where does the circle x^2 + 3x + y^2 + 6y - 4 = 0 intersect the y-axis.

1 Answer | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

At the points where the circle intersects the y-axis, the x-coordinate is equal to 0. To determine the y-intercepts the equation y^2 + 6y - 4 = 0 has to be solved.

y^2 + 6y - 4 = 0

y1 = `(-6 + sqrt(42))/2` and y2 = `(-6 - sqrt(42))/2`

The circle x^2 + 3x + y^2 + 6y - 4 = 0 intersects the y-axis at `(0, (-6 + sqrt(42))/2)` and `(0, (-6 - sqrt(42))/2)` .

We’ve answered 333,763 questions. We can answer yours, too.

Ask a question