X2 + y2 - 14x + 10y + 65 = 0
will lie ENTIRELY in Quadrant
1 answer:
Rearranging to standard form
= (x - 7)^2 + (y + 5)^2 = -65 + 49 + 25
(x-7)^2 + (y+5)^2 = 9
so the centre is at (7,-5) and the radius is 3
so the circle will be entirely in Quadrant 4.
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