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 The locus of the mid-points of the chords of the circle \mathrm{x^{2}+y^{2}-2 x-6 y-10=0} which pass through the origin is

Option: 1

\mathrm{x^{2}+y^{2}+x-3 y=0}


Option: 2

\mathrm{\quad x^{2}+y^{2}-x-3 y=0}


Option: 3

\mathrm{x^{2}+y^{2}-x+3 y=6}


Option: 4

\mathrm{x^{2}+y^{2}-x-3 y=6}


Answers (1)

best_answer

 Let \mathrm{(h, k)} be the coordinates of the mid point.

Equation of the chord whose mid point is \mathrm{(h, k)} is

\begin{aligned} & x h+y k-(x+h)-3(y+k)-10=h^{2}+k^{2}-2 h-6 k-10 \\ & \text { (using } \left.T=S_{1}\right) \\ & \text { i.e., } h^{2}+k^{2}-h(x+2)-k(6+y)+x+h+3 y+3 k=0 \end{aligned}

This chord passes through the origin (0,0)
Hence \mathrm{h^{2}+k^{2}-h-3 k=0}
Hence the required locus is \mathrm{x^{2}+y^{2}-x-3 y=0}.

Alternative:
If \mathrm{m_{1}} is the slope of the chord, then \mathrm{m_{1}=\frac{k-0}{h-0}}
If \mathrm{m_{2}} is the slope of the perpendicular to the chord then \mathrm{m_{2}=\frac{k-3}{h-1}} is the centre of the given circle. But
\mathrm{m_{1} m_{2}=-1 \Rightarrow h^{2}+k^{2}-\mathrm{h}-3 \mathrm{k}=0}
Hence the required locus is \mathrm{x^{2}+y^{2}-x-3 y=0}.

Hence (B) is the correct answer.

 

Posted by

seema garhwal

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