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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Apr 2007 12:06:33 IST
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1. The locus of the centre of the circle which touches the centre of the circle x2+ y2-6x-6y+14=0 and also touches the y-axis.
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21 Apr 2007 12:20:15 IST
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let the equation of the circle be (x-h)2+(y-k)2=r2 since it touches the y axis put x=0 in the above equation and put D=0 you will get, r2=h2 Now, the circle also passes through the point (3,3) so, (h-3)2+(k-3)2=h2 so, the locus is (y-3)2=6(x-3/2)
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21 Apr 2007 19:09:15 IST
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good work joyfrancis.
you can also make condition by using the fact that:
distance between (h,k) & (3,3) = radius= perpendicular distance from (h,k) to y axis
or (h-3)2+(k-3)2 = (radius)2 = h2
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21 Apr 2007 20:26:54 IST
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thank you sir
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21 Apr 2007 21:24:31 IST
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thank u joyfrancis
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