JEE Main 2006Annual, Paper 1MathematicsEasyCoordinate Geometry478× since 2002If the lines 3x−4y−7=03x - 4y - 7 = 03x−4y−7=0 and 2x−3y−5=02x - 3y - 5 = 02x−3y−5=0 are two diameters of a circle of area 49π49\pi49π square units, the equation of the circle is :AOption A. x2+y2+2x −2y−47=0 \,{x^2} + {y^2} + 2x\, - 2y - 47 = 0\,x2+y2+2x−2y−47=0BOption B. x2+y2+2x −2y−62=0 \,{x^2} + {y^2} + 2x\, - 2y - 62 = 0\,x2+y2+2x−2y−62=0COption C. x2+y2−2x +2y−62=0{x^2} + {y^2} - 2x\, + 2y - 62 = 0x2+y2−2x+2y−62=0DOption D. x2+y2−2x +2y−47=0{x^2} + {y^2} - 2x\, + 2y - 47 = 0x2+y2−2x+2y−47=0CheckSkip · show solution
