Thepointof intersectionofthelinex+3y=0and2x−7y=0is(0,0)Let(h,k)bethecentreofcirclewithradiusa.thus,itsequationwillbe(x−h)2+(y−k)2=a2Thepointoftheintersectionofthelinex+y+1=0andx−2y+4=0is(−2,1)∴h=−2,k=1∴Equationoftherequiredcircle=(x+2)2+(y−1)2=a2−−−−(1)also,equation(1)passesthrough(0,0)∴(0+2)2+(0−1)2=a24+1=a2∴a=√5(∴a>0)substitutingthevaluesfoainequation(1)(x+2)2+(y−1)2=5x2+4+4x+y2+1−2y=5x2+4x+y2−2y=0Hence,Itistherequiredanswer.