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

If the lines $\dpi{100} 3x-4y-7=0\; and\; 2x-3y-5=0\;$  are two diameters of a circle of area $\dpi{100} 49\pi$  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)

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

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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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