Circumference of circle =10π 2πr=10π (let r be the radius of circle) r = 5 Now, given that 2 diameters of circle lies along 2x+3y+1=0 and 3x−y−4=0 ∴ Centre is point of intersection of diameters 2x+3y+1=0 ...(i) 3x−y−4=0 ⇒y=3x−4 Put y=3x−4 is Eq. (i), we get 2x+3(3x−4)+1=0 ⇒2x+9x−12+1=0 11x−11=0,x=1,y=3−4=−1 ∴ Center is (1,−1) Now, equation of circle whose centre is (1,−1) and radius is 5 [Equation of circle-centre (x1,y1), radius =r(x−x1)2+(y−y1)2=r2] ⇒(x−1)2+(y+1)2=52 ⇒x2−2x+1+y2+2y+1=25 ⇒x2+y2−2x+2y−23=0