How do you find the graph of the equation $x^2 - 6x + y^2 + 2y + 6 = 0$?
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8hint: complete the squares. this represents a circle. – abel Jul 13 '15 at 21:35
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possible duplicate of Complete the squares to find the center and radius of the circle – Jul 14 '15 at 03:18
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You try to reduce the LHS into a perfect square.
$$x^2- 6x = x^2 - 6x + 9 - 9 = (x-3)^2 - 9$$
$$y^2 + 2y + 6 = y^2 + 2y + 1 + 5 = (y+ 1)^2 + 5$$
After putting these back in the equation, and re-arranging, you'll have the form:
$$(x-a)^2 + (y - b)^2 = r^2$$
This is a circle of center $(a,b)$ and radius $r$.
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The left hand side of the equation can be simplified by completing the square. $$x^2-6x+y^2+2y+6 $$ $$=x^2-6x+9-9+y^2+2y+1-1+6=0$$
Because $x^2-6x+9 = (x-3)^2$ and $y^2+2y+1 = (y+1)^2$ we obtain: $$=(x-3)^2-9+(y+1)^2-1+6=0$$ $$\implies(x-3)^2 +(y+1)^2 -4 = 0$$ $$\implies(x-3)^2 +(y+1)^2 = 4 $$ The equation of a circle with center at the point $(h, k)$ and radius $r$ is: $$(x-h)^2+(y-k)^2 = r^2$$ Comparing these equations we get that the center of the circle is $(3,-1)$ and the radius is $2$ and this is very easy to graph.
Rick
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