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Circle Co-ordinate Geometry Help (1 Viewer)

kmlb123

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Nov 9, 2008
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hi everyone, thanks for taking your time in reading this, but I've been stuck on this question for a bit and just wondering what i'm doing terribly wrong.

Write down the coordinates of the centre, and the length of the radius of the following circle:

a) 2x^2+2y^2-8x+5y+3=0

My working out was:

1) 2(x^2-4x+y^2+2.5)= -3
2) x^2-4x+y^2+2.5= -1.5
3) (x^2-4x+4)+(y^2+2.5+1.5626)= -1.5-4-1.5625
4) (x-2)^2+(y-1.25)^2= -7.0625

Therefore, centre is (2,-1.25), but then i realised that r^2 was negative and you could not really square root it to find the radius of the circle. The answers say the answer for the radius is 1/465. Any ideas?
 

Timothy.Siu

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2009
a) 2x^2+2y^2-8x+5y+3=0

x^2+y^2-4x+5/2y+3/2=0
(x-2)^2-4+(y+5/4)^2-25/16+3/2=0 (completing the square)
(x-2)^2+(y+5/4)^2-4.0625=0
(x-2)^2+(y+5/4)^2=4.0625

centre: (2,-5/4)
radius: root(65/16)=1/4(root65)
 

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