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Help me please, - Co-ordinate geometry - JEE Main-3

If the lines 3x-4y-7=0\; and\; 2x-3y-5=0\;  are two diameters of a circle of area 49\pi  square units, then the equation of the circle is

  • Option 1)

    x^{2}+y^{2}+2x-2y-47=0

  • Option 2)

    x^{2}+y^{2}+2x-2y-62=0

  • Option 3)

    x^{2}+y^{2}-2x+2y-62=0

  • Option 4)

    x^{2}+y^{2}-2x+2y-47=0

 
Answers (1)
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As we learnt in 

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}

 

 Point of intersection of 3x - 4y - 7= 0 and 2x - 2y - 5 = 0 is (1, -1) which is the center and \pi r^{2}=49 \pi

r = 7

Equation is x2 + y2 - 2x + 2y - 47 = 0.


Option 1)

x^{2}+y^{2}+2x-2y-47=0

this is incorrect

Option 2)

x^{2}+y^{2}+2x-2y-62=0

this is incorrect

Option 3)

x^{2}+y^{2}-2x+2y-62=0

this is incorrect

Option 4)

x^{2}+y^{2}-2x+2y-47=0

this is correct

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