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    				| 1. | From the focus of parabola `y^2 = 8x` as centre, a circle is described so that a common chord is equidistant from vertex and focus of the parabola. The equation of the circle is | 
| Answer» OP=PF OP=OF/2=1 `y^2=8x` `y^2=8` `y=pm2sqrt2` `FN=sqrt((2-1)^2+(2sqrt2-0)^2)=sqrt9=3` Circle=`(x-2)^2+y^2=9`. | |