The equation of the circle whose one diameter is PQ, where the ordinates of P, Q are the roots of the equation x2+2x−3=0 and the abscissae are the roots of the equation y2+4y−12=0,is

Let Px1,y1 and Qx2,y2 be the end points of diameter PQ. Thenx1+x2=−2,x1x2=−3,y1+y2=−4 and y1y/2=−12The equation of the circle having PQ as a diameter isx2+y2−xx1+x2−yy1+y2+x1x2+y1y2=0x2+y2+2x+4y−3−12=0x2+y2+2x+4y−15=0