Point of intersection of 3x−4y−7=0 and 2x−3y−5=0 is (1,−1), which is the centre of the circle and radius =7. ∴ Equation is (x−1)2+(y+1)2=49⇒x2+y2−2x+2y−47=0
If the lines 3x−4y−7=0 and 2x−3y−5=0 are two diameters of a circle of area 49π square units, the equation of the circle is
Held on 30 Apr 2006 · Verified 6 Jul 2026.
x2+y2+2x−2y−47=0
x2+y2+2x−2y−62=0
x2+y2−2x+2y−62=0
x2+y2−2x+2y−47=0
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