Let the abscissae of the two points P and Q be the roots of 2x2−rx+p=0, and the ordinates of P and Q be the roots of x2−sx−q=0. If the equation of the circle described on PQ as diameter is 2(x2+y2)−11x−14y−22=0, then 2r+s−2q+p is equal to:
Answer: 7
Solutions · 1
Official solution
x1+x2=2r,x1x2=2p; y1+y2=s,y1y2=−q. Circle on diameter PQ:
\[x^2+y^2-\frac r2x-sy+\left(\frac p2-q\right)=0.\]
Comparing with x2+y2−5.5x−7y−11=0: r=11,s=7,2p−q=−11⇒p−2q=−22.
2r+s−2q+p=22+7+(p−2q)=29−22=7.