EAMCET Engineering 2008 Solved Paper

Since, the lines 2x−3y=5 and 3x−4y=7 are the diameters of a circle. Therefore, the point of intersection is the centre of the circle. On solving the given equations, we get x=1 and y=−1ie, the centre of the circle.
∴ Required equation of circle is
‌(x−1)2+(y+1)2=72
‌⇒‌‌x2+y2−2x+2y+2=49
‌⇒‌‌x2+y2−2x+2y−47=0
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