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:
Solutions · 1
Official solution
Point of intersection of 3x−4y−7=0 and
2x−3y−5=0 is \[\left( {1, - 1} \right)\] which is the center of the
circle and radius =7
\[\therefore\] Equation is \[{\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 49\]
\[ \Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0\]