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  • Can someone explain - A square is inscribed in the circle with its sides parallel to the coordinate axes. Then the dist - Co-ordinate geometry - JEE Main

 

A square is inscribed in the circle x^{2}+y^{2}-6x+8y-103=0 with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is :

  • Option 1)

     

    \sqrt{41}

  • Option 2)

    13

  • Option 3)

     

    \sqrt{137}

  • Option 4)

     

    6

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}

 

Radius of circle, R=\sqrt{r^{2}+f^{2}-c}

R=\sqrt{9+16+103}

=\sqrt{128}=8\sqrt{2}

circle equation can be written as -

(x-3)^{2}+(y+4)^{2}=(8\sqrt{2})^{2}

 

 

OC=\sqrt{11^{2}+42}=\sqrt{137}

OO=\sqrt{25+16}=\sqrt{41}

Nearest vertex from origin is \sqrt{41}

 


Option 1)

 

\sqrt{41}

Option 2)

13

Option 3)

 

\sqrt{137}

Option 4)

 

6

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