Let the abscissae of the two points P and Q on a circle be the roots of x2−4x−6=0 and the ordinates of P and Q be the roots of y2+2y−7=0. If PQ is a diameter of the circle x2+y2+2ax+2by+c=0, then the value of (a+b−c) is
(A)12✓ Correct
(B)13
(C)14
(D)16
Answer: (A)
Solution
Since PQ is a diameter, the equation of the circle in diameter form is (x−x1)(x−x2)+(y−y1)(y−y2)=0, where x1,x2 are roots of x2−4x−6=0 and y1,y2 are roots of y2+2y−7=0. Expanding: x2−(x1+x2)x+x1x2+y2−(y1+y2)y+y1y2=0. By Vieta's formulas: x1+x2=4, x1x2=−6, y1+y2=−2, y1y2=−7. Thus the circle is x2+y2−4x+2y−13=0. Comparing with x2+y2+2ax+2by+c=0, we get 2a=−4 so a=−2, 2b=2 so b=1, and c=−13. Therefore a+b−c=−2+1−(−13)=12.
JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option
