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Question 405072: put in standard form, state the type of graph it will produce, and center,key point, and/or radius for the following:
X^2 + (y-2)^2=7
x^2+6x+4y^2+8y=-13 ( for this one i know one must complete the square, but the 4
in front of the y throws me off on the equation into standard form)
xy=7 ( im pretty sure that this equation doesn't make a graph, but im not confident about my answer on noting)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! put in standard form, state the type of graph it will produce, and center,key point, and/or radius for the following:
X^2 + (y-2)^2=7
x^2+6x+4y^2+8y=-13 ( for this one i know one must complete the square, but the 4
in front of the y throws me off on the equation into standard form)
..
x^2+(y-2)^2=7
This is a circle with center at (0,2) with radius=sqrt(7)
Standard form for circle:(x-h)^2+(y-k)^2=r^2.
In this case, h=0,k=2, and r^2=7
..
x^2+6x+4y^2+8y=-13
complete the square
(x^2+6x+9)+4(y^2+2y+1)=-13+9+4=0
(x+3)^2+4(y+1)^2=0
This does not fit into any of the equations for conics that I know of. Maybe there is a typing error here. If you try to solve it for y, you will get imaginary solutions.
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