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
A
x2+y2+2x−4y−47=0
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B
x2+y2=49
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C
x2+y2−2x+2y−47=0
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D
x2+y2=17
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Solution
The correct option is Dx2+y2−2x+2y−47=0 2x−3y=5−−−−(1) 3x−4y=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 ⇒x2+y2−2x+2y−47=0