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When graphed, the circle with equation

x2 + y2 - 14x + 10y + 65 = 0

will lie ENTIRELY in Quadrant

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Recall what each quadrant means.
Quadrant I (x, y)
Quadrant II (-x, y)
Quadrant III (-x, -y)
Quadrant IV (x, -y)

Now, we just need to complete the square twice to produce the general form of a circle.

x^2 - 14x + 49 + y^2 + 10y + 25 + 65 - 49 - 25 = 0
(x - 7)^2 + (y + 5)^2 - 9 = 0
(x - 7)^2 + (y + 5)^2 = 9

Centre at (7, -5)
Always lies in Quadrant IV
User Ben Hillier
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