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the locus of foot of the perpendicular drawn from the origin to a variable line passing through a fixed point is circle whose diameter is ?
 
Ans is [square root of 13 ] HOW.???
 
 

sreya r , 6 Years ago
Grade 12
anser 1 Answers
Arun

Last Activity: 6 Years ago

Dear student
 
you have not define the fixed point.
I am assuming it as (a, b)
Let the line passing through (a,b) be,
y= mx-am+b,          ........(i)
Now a line perpendicular to this passing through origin is,
y=-x/m                  .......(ii)
let the feet of the perpendicular be (h,k)
From (i)
k= mh-am+b           .......(iii)
From (ii)
m= -h/k,
Replacing this in (iii)
k= (-h/k)*h + a(h/k) +b,
hence, k2=-h2+ah+bk,
So h2+k2= ah+bk,
Hence the locus is x2+y2= ax+by

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