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  • The equation of line joining midpoints of the tangents   drawn from or

The equation of line joining midpoints of the tangents \mathrm{O P, O Q}  drawn from origin O to the circle\mathrm{x^2+y^2-4 x-6 y+8=0}  is
 

Option: 1

\mathrm{3 x+2 y-4=0}


Option: 2

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


Option: 3

\mathrm{3 x-2 y+4=0}


Option: 4

\mathrm{2 x+3 y-4=0}


Answers (1)

best_answer

radical axis of \mathrm{x^2+y^2-4 x-6 y+8=0}  and \mathrm{x^2+y^2=0} is \mathrm{2 x+3 y-4=0}

Method II : equation of chord of contact (PQ) of point ' O ' w.r.t. given circle is

\mathrm{x(0)+y(0)-2(x+0)-3(y+0)+8=0 \Rightarrow 2 x+3 y-8=0 }

Point \mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right), \mathrm{Q}\left(\mathrm{x}_2, \mathrm{y}_2\right) lie on above line
Hence 

\mathrm{ 2 x_1+3 y_1-8=0, \quad 2 x_2+3 y_2-8=0 }
\mathrm{ 2\left(\frac{x_1}{2}\right)+3\left(\frac{y_1}{2}\right)-4=0, \quad 2\left(\frac{x_2}{2}\right)+3\left(\frac{y_2}{2}\right)-4=0}

Hence equation of line joining mid point is \mathrm{ 2 x+3 y-4=0}

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