AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

since it is given that abscissae of two points P and Q are roots of the equation 2x2+4x−7=0 and their coordinates are the roots of the equation 3x2−12x−1=0
Let P(x1,y1) and Q(x2,y2) then,
x1+x2=−2 and y1+y2=4
So, the center of circle with PQ as diameter is,
(‌

x1+x2

2

,‌

y1+y2

2

)=(−1,2)