
Given that the roots of x2−4x−6=0 are the abscissa of the end of diameter
i.e. x1+x2=4,x1x2=−6
and roots of y2+2y−7=0 are ordinate of the end of diameter
i.e. y1+y2=−2,y1y2=−7
Now, equation of the circle will be
(x−x1)(x−x2)+(y−y1)(y−y2)=0
i.e. x2−(x1+x2)x+x1x2+y2−(y1+y2)y+y1y2=0
⇒x2+y2−4x+2y−13=0
∴a=−2,b=1,c=−13
⇒a+b−c=−2+1+13=12