﻿ Let S ≡ x2 + y2 + 2gx + 2fy + c = 0 be a given circle. find the locus of the foot of the perpendicular drawn from the origin upon any chord of S which subtends a right angle at the origin. : Kaysons Education

# Let S ≡ x2 + y2 + 2gx + 2fy + c = 0 Be A Given Circle. Find The Locus Of The Foot Of The Perpendicular Drawn From The Origin Upon Any Chord Of S which Subtends A Right Angle At The Origin.

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## Question

### Solution

Correct option is

Equation of the given circle is

S = x2 + y2 + 2gx + 2fy + c = 0              …(1)

Any chord of the circle be

lx + my = 1                                            …(2)

Chord (2) subtends a right angle at the origin  So making equation (1) homogeneous with the help of (2), we get

(x2 + y2) + (2gx + 2fy) (lx + my) + c(lx + my)2 = 0

x2 + y2 + 2glx2 + 2gmxy + 2flxy + 2fmy2 + cl2x2 + cm2y2 + 2clmxy = 0

Apply the condition of perpendicularity

We get

Any line through origin ⊥ to (1) is

mx – ly = 0            …(4)

Both (2) and (4) give the foot of perpendicular whose locus is obtained by eliminating the variable lm between (2), (3) and (4).

On solving (2) & (4)

We get,

Putting in (2) we get

Putting the value of l & m in  (3) we get

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