Find the locus of the foot of the perpendicular drawn from the origin to any chord of the circle S=x^2+y^2+2gx+2fy+c=0 which subtends a right angle at the origin.
Answers
Answer:
Let S=x
2
+y
2
+2gx+2fy+c=0 be a given circle . Then the locus of the foot of the perpendicular drawn from the origin upon any chords of S which subtends right angle at the origin is:
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Given circle =x
2
+y
2
+2gx+2fy+c=0
Let LM be the chord
of circle C and subtend right angle at (0,0)O,P(h,k) be fool of ⊥ from O on chord LM
as OP⊥ML, eqn of LM is:
y−k=
k
h
(x−h)
hx+ky=h
2
+k
2
.......... (1)
equation of pair of line joining point of intersection of (1) with S=0⇒x
2
+y
2
+2(gx+fy)
h
2
+k
2
(hx+ky
+c
(h
2
+k
2
)
2
(hx+ky)
2
=0 ........... (2)
as lines in (2) are at right angles
1+1=
h
2
+k
2
2gh+2fk
+c
(h
2
+k
2
)
(h
2
+k
2
)
=0
h
2
+k
2
+gh+fk+
2
c
=0
∴ Locus of (h,k) is ⇒S=x
2
+y
2
+gx+fy+
2
c
=0
Option (A)