Find the equation of the circle which touches the line y=2 ,passes through origin and the point where the curve y^2-2x+8=0 meets the x-axis.

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Let equation of cirlce be x2+y2+2gx+2fy+c=0 whose centre is -g,-fand radiusr=g2+f2-cAs circle passes through 0,0, hence0+0+0+0+c=0c=0 x2+y2+2gx+2fy=0 ;iCurve y2-2x+8=0 meets x axis where y=0-2x+8=0x=4Hence cirlce passes through 4,0 .Put 4,0 in i16+0+8g+0=0g=-2Hence i becomes: x2+y2-4x+2fy=0 ;iiCentre=-g,-f=2,-fr=g2+f2-c=4+f2As cirlce touches y-2=0, hence length of perpendicular from centre to line y-2=0 will be equal to radius of cirlce.0-f-202+12=4+f2-f+2=4+f2f+2=4+f2Squarring both sides, we get:f+22=4+f2f2+4+4f=4+f2f=0Put this in iix2+y2-4x+0=0Hence equation of required cirlce isx2+y2-4x=0
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