If the lines 2x+3y+1=0\; and\; 3x-y-4=0  lie along diameters of a circle of circumference 10\pi , then the equation of the circle is

  • Option 1)

    x^{2}+y^{2}+2x+2y-23=0

  • Option 2)

    x^{2}+y^{2}-2x-2y-23=0

  • Option 3)

    x^{2}+y^{2}-2x+2y-23=0

  • Option 4)

    x^{2}+y^{2}+2x-2y-23=0

 

Answers (1)

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}

 

 coordinates of centre is the point of  intersection of 2x+3y + 1 =0 and 3x - y - 4 = 0

    2x+3y=-1

    \underline{9x-3y=12}

     11x=11\Rightarrow x=1

    2(1)+3y=-1

    y=-1

Point\ is (1,-1)

also, 2\pi r=10\pi

\Rightarrow r=5

Equation is 

(x-1)^{2}+(y+1)^{2}=5^{2}\\ x^{2}+y^{2}-2x+2y-23=0


Option 1)

x^{2}+y^{2}+2x+2y-23=0

This option is incorrect 

Option 2)

x^{2}+y^{2}-2x-2y-23=0

This option is incorrect 

Option 3)

x^{2}+y^{2}-2x+2y-23=0

This option is correct 

Option 4)

x^{2}+y^{2}+2x-2y-23=0

This option is incorrect 

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