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Question

Find the equation of the circle passing through the points of intersection of the line x+3y=0 and 2x7y=0 and whose centre is the point of intersection of the lines x+y+1=0 and x2y+4=0.

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Solution

Thepointof intersectionofthelinex+3y=0and2x7y=0is(0,0)Let(h,k)bethecentreofcirclewithradiusa.thus,itsequationwillbe(xh)2+(yk)2=a2Thepointoftheintersectionofthelinex+y+1=0andx2y+4=0is(2,1)h=2,k=1Equationoftherequiredcircle=(x+2)2+(y1)2=a2(1)also,equation(1)passesthrough(0,0)(0+2)2+(01)2=a24+1=a2a=5(a>0)substitutingthevaluesfoainequation(1)(x+2)2+(y1)2=5x2+4+4x+y2+12y=5x2+4x+y22y=0Hence,Itistherequiredanswer.

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