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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....

3 years ago

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Answers : (2)

                                        x^2+y^2=x
                                        
3 years ago
                                        

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)


y=sinØ/2;     (2)


from (1) and (2), (cosØ)^2= (2x-1)^2,   (3)


                          (sinØ)^2=(2y)^2;     (4)


adding (3) and (4)


     sqr(2x-1)+sqr(2y)=1,


 or x^2 +y^2-x=0 ,which is the required eq.


 

3 years ago

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