lf the lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are two diameters of a circle of radius $7$ then the equation of the circle is

x^{2} + y^{2} + 2 x - 4 y - 47 = 0

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x^{2} + y^{2} = 49

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

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x^{2} + y^{2} = 17

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Solution

The correct option is D $x^{2} + y^{2} - 2 x + 2 y - 47 = 0$
$2 x - 3 y = 5 - - - - (1)$
$3 x - 4 y = 7 - - - - (2)$
Intersection of this lines given centre of circle
$∴$ from (1) & (2),
$n = 1, y = - 1$
So, equation of circle is
$(x - 1)^{2} + (y + 1)^{2} = (7)^{2}$
$\Rightarrow x^{2} + y^{2} - 2 x + 2 y - 47 = 0$

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lf the lines 2x−3y=5 and 3x−4y=7 are two diameters of a circle of radius 7 then the equation of the circle is

The correct option is D x2+y2−2x+2y−47=02x−3y=5−−−−(1)3x−4y=7−−−−(2)Intersection of this lines given centre of circle∴ from (1) & (2),n=1,y=−1So, equation of circle is(x−1)2+(y+1)2=(7)2⇒x2+y2−2x+2y−47=0

lf the lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are two diameters of a circle of radius $7$ then the equation of the circle is