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  • The locus of the foot of the perpendicular from the origin to chords of the circle <img alt="\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-4 \mathrm{y}-4=0" src="https://entrancecorner.oncodecogs

The locus of the foot of the perpendicular from the origin to chords of the circle \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-4 \mathrm{y}-4=0  which subtend a right angle at the origin is

Option: 1

\mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{x}-2 \mathrm{y}-2=0


Option: 2

\mathrm{2\left(x^{2}+y^{2}\right)-2 x-4 y+3=0}


Option: 3

\mathrm{x^{2}+y^{2}-2 x-4 y+4=0}


Option: 4

\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{x}+2 \mathrm{y}-2=0


Answers (1)

best_answer

Equation to the chord is \mathrm{xx}_{1}+\mathrm{yy}_{1}=x_{1}^{2}+y_{1}^{2}

Where M\left(x_{1}, y_{1}\right) is a foot of \perpor from the origin homogenize the equation to the circle given, as we get
\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)\left(x_{1}^{2}+y_{1}^{2}\right)^{2}-(2 \mathrm{x}+4 \mathrm{y})\left(\mathrm{xx}_{1}+\mathrm{yy}_{1}\right) \cdot\left(x_{1}^{2}+y_{1}^{2}\right)-4\left(\mathrm{xx}_{1}+\mathrm{yy}_{1}\right)^{2}= 0



This represents a pair of perpendicular lines through the origin i.e.
2\left(x_{1}^{2}+y_{1}^{2}\right)^{2}-\left(x_{1}^{2}+y_{1}^{2}\right)\left(2 x_{1}+4 y_{1}\right)-4\left(x_{1}^{2}+y_{1}^{2}\right)=0

\text{giving the locus of }\, \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\, is
2\left(x^{2}+y^{2}\right)-2 x-4 y-4=0
i.e. \, x^{2}+y^{2}-x-2 y-2=0

 

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