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A variable line lx+my=1 (where l and m are parameters) intersect a circle x^(2)+y^(2)-4x+3=0 at the points P and Q. The chord PQ subtends a right angle at the origin. If the locus of foot of perpendicular drawn from origin on the given variable lines is lambdax^(2)+muy^(2)-4x+3=0, then find the value of (lambda+mu). |
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Answer» Solution :Homogenising circle with variable line `x^(2)+y^(2)-4x(lx+my)+3(lx+my)^(2)=0` SINCE OP and OQ are at right angle. `:."Coefficient of "x^(2)+"coeff. Of "y^(2)=0` `(1-4l+3l^(2))+(1+3m^(2))=0` `3l^(2)+3m^(2)-4l+2=0""…(1)` Let FOOT of perpendicular is (h,k) `:.""lh+mk=1""...(2)` Equation of OM is `y=(m)/(l)x` `mh=lk""...(3)` From (2) and (3) we get `l=(h)/(h^(2)+k^(2))" and "m=(k)/(h^(2)+k^(2))` `l^(2)+m^(2)=(1)/(h^(2)+k^(2))" using this in (1)"` `(3)/(h^(2)+k^(2))-(4h)/(h^(2)+k^(2))+2=0` `:."""Required LOCUS is "2x^(2)+2y^(2)-4x+3=0` `:.""lambda=mu=2` `:.""lambda+mu=4` 
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