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Let \mathrm{S \equiv x^2+y^2+2 g x+2 f y+c=0} be a given circle. Find the locus of the foot of perpendicular drawn from origin upon any chord of S which subtends a right angle at the origin.

 

Option: 1

\mathrm{2\left(x^2+y^2\right)+(g x+f y)+c=0}


Option: 2

\mathrm{2\left(x^2+y^2\right)+2(g x+f y)+c=0}


Option: 3

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


Option: 4

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


Answers (1)

best_answer

The slope of perpendicular drawn from origin O to the chord \mathrm{AB}=\frac{k}{h}

where M(h, k) is the foot of perpendicular.

Slope of \mathrm{AB}=\frac{h}{-k}. Equation of AB is then

\mathrm{\begin{gathered} \frac{h}{k}(\mathrm{x}-\mathrm{h}) \\ \mathrm{y}-\mathrm{k}=-\frac{1}{\mathrm{x}}(\mathrm{k}) \\ \text { or, } \mathrm{hx}+\mathrm{ky}=\mathrm{h}^2+\mathrm{k}^2 \end{gathered}} -------(1)

By standard result, the combined equation of OA and OB is obtained by making the equation of circle homo-geneous with the help of (1).

\mathrm{ \mathrm{x}^2+\mathrm{y}^2+(2 \mathrm{gx}+2 \mathrm{fy})\left(\frac{h x+k y}{h^2+k^2}\right)+\mathrm{c}\left(\frac{\mathrm{hx}+\mathrm{ky}}{\mathrm{h}^2+\mathrm{k}^2}\right)^2=0 }

Since the angle between OA and OB is 90°, the sum of the coefficients of \mathrm{x^{2} } and \mathrm{y^{2} }  in above equation is zero.

\mathrm{\begin{aligned} & {\left[1+\frac{2 g h}{h^2+k^2}+\frac{c h^2}{\left(h^2+k^2\right)^2}\right]+\left[1+\frac{2 f k}{h^2+k^2}+\frac{c k^2}{\left(h^2+k^2\right)^2}\right]=0} \\ & \text { or, } 2+\frac{2(g h+f k)}{h^2+k^2}+\frac{c}{h^2+k^2}=0 \end{aligned}}

The locus of (h, k) is therefore  \mathrm{2\left(x^2+y^2\right)+2(g x+f y)+c=0}

Posted by

HARSH KANKARIA

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