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

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jimone | Student | eNoter

Posted June 29, 2012 at 4:59 PM via web

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

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 29, 2012 at 5:04 PM (Answer #1)

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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)` .

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