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The lines 2x-3y=5\; and\; 3x-4y=7   are diameters of a circle having area as 154 sq. units. Then the equation of the circle is

  • Option 1)

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

  • Option 2)

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

  • Option 3)

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

  • Option 4)

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

 

Answers (1)

As we learnt in

Equation of a circle -

\left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}

- wherein

Circle with centre \left ( h,k \right ) and radius r.

 

 Centre is a point of intersection of

2x-3y=5 \ and \ 3x -4y=7

i.e. \ x=1, y=-1

also \pi r^{2}=154

\frac{22}{7}\times r^{2}=154

\Rightarrow r^{2}=49

\Rightarrow r=7

equation of circle is

(x-1)^{2}+(y-1)^{2}=7^{2}

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


Option 1)

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

This option is incorrect.

Option 2)

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

This option is correct.

Option 3)

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

This option is incorrect.

Option 4)

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

This option is incorrect.

Posted by

Sabhrant Ambastha

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