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  • Let <img alt="\mathrm{S=x^2+y^2+2 g x+2 f y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BS%3Dx%5E2&plus;y%5E2&plus;2%20g%20x&plus;2%20f%20y&plus;c%3D0

Let \mathrm{S=x^2+y^2+2 g x+2 f y+c=0}  be a given circle. 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 is

Option: 1

\mathrm{x^2+y^2+g x+f y+\frac{c}{4}=0}


Option: 2

\mathrm{x^2+y^2+2 g x+f y+\frac{c}{3}=0}


Option: 3

\mathrm{x^2+y^2+g x+f y+c=0}


Option: 4

\mathrm{x^2+y^2+g x+f y+\frac{c}{2}=0}


Answers (1)

best_answer

Let lx + my = 1 ...(1)
be a chord of the circle

\mathrm{x^2+y^2+2 g x+2 f y+c=0}      ...[A]

It subtends a right angle at the origin.
∴ The lines joining the origin to the points of intersection of chord and circle are at right angles. Making (A) homogeneous with the help of (1).

\mathrm{\therefore x^2+y^2+(2 g x+2 f y)(l x+m y)+c(l x+m y)^2=0}     ...[2]

A line through the origin and perpendicular to (1) is

mx – ly = 0 ...(3)

\mathrm{\text { or } \frac{x}{l}=\frac{y}{m}=\frac{x^2+y^2}{l x+m y}=\frac{x^2+y^2}{1}}   ....[4]

\mathrm{\Rightarrow l=\frac{x}{x^2+y^2}, m=\frac{y}{x^2+y^2}}

Substituting values of l and m in (2), we get

\mathrm{\begin{aligned} & c \frac{1}{x^2+y^2}+\frac{2(g x+f y)}{x^2+y^2}+2=0 \\ & \text { or } x^2+y^2+g x+f y+\frac{c}{2}=0 . \end{aligned}}

 

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Nehul

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