Free JEE ADV-13 Course
from the origin,chords are drawn to the circle square(x-1) +square(y)=1.the equation of the locus of the mid pt. of these chord is....
O(0,0) lies on the circle.
general form of any point on this circle is given by (-g+r(cosØ),-f+r(sinØ)) hence here it is P(1+1(cosØ),1(sinØ)).
thus for mid pt. of chord OP:
x= (0+1+1cosØ)/2 =>(1+cosØ)/2 (1)
from (1) and (2), (cosØ)^2= (2x-1)^2, (3)
adding (3) and (4)
or x^2 +y^2-x=0 ,which is the required eq.