NVAMediumJEE Main 2026 · 21 January, Shift 2Circle Equation & Properties

Mathematics Question from JEE Main 2026 · 21 January, Shift 2

If PP is a point on the circle x2+y2=4x^2 + y^2 = 4, QQ is a point on the straight line 5x+y+2=05x + y + 2 = 0 and xy+1=0x - y + 1 = 0 is the perpendicular bisector of PQPQ, then 1313 times the sum of abscissa of all such points PP is _____.

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Practice more Circle Equation & Properties questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Related questions