Given, equation of circles are 2x−3y+12=0...(i) x+4y−5=0....(ii) On solving Eq (i) and Eq (ii). we get x=−3 and y=2 Hence coordinate of centre is(−3,2) Given area of circle = 154 sq units ⇒πr2=154   [where r is radius of circle] ⇒
22
7
×r2=154 r2=49 ⇒r=7 Here centre is (−32) and radius =7 Then, required of equation of circle is {x−(−3)}2+(y−2)2=(7)2 ⇒(x+3)2+(y−2)2=49 ⇒x2+9+6x+y2+4−4y−49 ⇒x2+y2+6x−4y+13−49=0 ⇒x2+y2+6x−4y−36=0