The locus of the foot of the perpendicular, from the origin to chords of the circle x 2 + y 2 – 4x – 6y – 3 = 0 which subtend a right angle at the origin, is -
Text Solution
Verified by ExpertsA
The equation to one such chord, on which the foot of the perpendicular from the origin is
(x 1 , y 1 ) is xx 1 + yy 1 = 
+ 
. Using this Homogenise x 2 + y 2 – 4x – 6y – 3 = 0 . This gives (x 2 + y 2 ) ( 
+ 
) 2 – 2(2x + 3y) (xx 1 + yy 1 ) (x 1 2 + y 1 2 ) – 3 (xx 1 + yy 1 ) 2 = 0. These two lines are at right angles.
∴ 2( 
+ 
) 2 – 2(2x 1 + 3y 1 ) ( 
+ 
) – 3 ( 
+ 
) = 0
+ 
≠ 0 and hence locus of
(x 1 , y 1 ) is 2(x 2 + y 2 ) – 2(2x + 3y) – 3 = 0.
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