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  • Solve it, If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is :

If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is : 

  • Option 1)

    \left ( x^{2}+y^{2} \right )^{2}=4Rx^{2}y^{2}

     

     

     

  • Option 2)

    \left ( x^{2}+y^{2} \right )^{2}=4R^{2}x^{2}y^{2}

  • Option 3)

    \left ( x^{2}+y^{2} \right )^{3}=4R^{2}x^{2}y^{2}

  • Option 4)

    \left ( x^{2}+y^{2} \right )\left ( x+y \right )=R^{2}xy

Answers (1)

best_answer

 

General form of a circle -

x^{2}+y^{2}+2gx+2fy+c= 0
 

- wherein

centre = \left ( -g,-f \right )

radius = \sqrt{g^{2}+f^{2}-c}

 

equation of line AB

y-k=\frac{h}{k}(x-h)\\\\hx+ky=h^{2}+k^{2}\\\\A\left ( \frac{h^{2}+k^{2}}{k},0 \right )\\\\B\left ( 0,\frac{h^{2}+k^{2}}{k} \right ) \\\\0\left ( 0,0 \right )\\\\AB=2R\\\\\ \frac{\left ( h^{2}+k^{2} \right )^{2}}{k^{2}}+\frac{\left ( h^{2}+k^{2} \right )^{2}}{h^{2}}=4R^{2}\\\\\Rightarrow \left ( h^{2}+k^{2} \right )\left ( \frac{h^{2}+k^{2}}{h^{2}k^{2}} \right )=4R^{2}\\\\\left ( x^{2}+y^{2} \right )^{3}=4R^{2}x^{2}y^{2}

 


Option 1)

\left ( x^{2}+y^{2} \right )^{2}=4Rx^{2}y^{2}

 

 

 

Option 2)

\left ( x^{2}+y^{2} \right )^{2}=4R^{2}x^{2}y^{2}

Option 3)

\left ( x^{2}+y^{2} \right )^{3}=4R^{2}x^{2}y^{2}

Option 4)

\left ( x^{2}+y^{2} \right )\left ( x+y \right )=R^{2}xy

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