Question
If P is a point on the circle x^ 2 +y^ 2 =4, Q is a point on the straight line 5 x+y+2=0 and x-y+1=0 is the perpendicular bisector of PQ, then 13 times the sum of abscissa of all such points P is \_\_\_\_.
JEE Main
Mathematics
Circle
2026
JEE Main 2026 (21 January Shift 2)
JEE Main Mathematics Question (2026) — Solution
Options
- A. A
- B. B
- C. C
- D. D
Answer
A. A
Step-by-step solution
Mid point of PQ lies on x - y + 1 = 0 2 + 2 - 2 - 5 - 2 2 + 1 = 0 2 + - 2 + 5 + 2 + 2 = 0 - + 3 + 2 = 0 ...(1) Slope of PQ is -1 2 + 5 + 2 2 - = -1 2 + 5 + 2 = -2 + + + 2 + 1 = 0 ...(2) eliminate from (1) and (2) + 5 = 1 , [0, 2 ] 5 2 2 2 = 2 ^2 2 2 = 0 = 1 or 2 = 5 = - 12 13 Sum of all possible values of abscissa of point P is = 2 1 + 2 ( -12 13 ) = 2 13 13 times sum of all possible values of abscissa of point P is 2.
Related: Mathematics — Circle · All PYQ Banks