JEE Main 2026 — Jan 21, Shift 2 — Mathematics solution

JEE Main2026Jan 21, Shift 2Mathematics

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 point P is _____.

Your answer

Solution

Parametrise $P = (2 cos θ, 2 sin θ)$ on the circle, and $Q = (α, - 5 α - 2)$ on the given line.

Condition 1 — midpoint of $P Q$ lies on $x - y + 1 = 0$ : $\frac{2 cos θ + α}{2} - \frac{2 sin θ - 5 α - 2}{2} + 1 = 0$ , simplifying to $cos θ - sin θ + 3 α + 2 = 0$ … (i).

Condition 2 — $P Q$ perpendicular to $x - y + 1 = 0$ (slope 1), so slope of $P Q = - 1$ : $\frac{2 sin θ + 5 α + 2}{2 cos θ - α} = - 1$ , giving $sin θ + cos θ + 2 α + 1 = 0$ … (ii).

Eliminate $α$ : from (i) $3 α = - cos θ + sin θ - 2$ , from (ii) $2 α = - sin θ - cos θ - 1$ . So $6 α = 2 (- cos θ + sin θ - 2) = 3 (- sin θ - cos θ - 1)$ . This reduces to $cos θ + 5 sin θ = 1$ .

Using half-angle: write $cos θ = 1 - 2 sin^{2} (θ /2)$ , $sin θ = 2 sin (θ /2) cos (θ /2)$ . Then $10 sin (θ /2) cos (θ /2) = 2 sin^{2} (θ /2)$ , factor: $sin (θ /2) [5 cos (θ /2) - sin (θ /2)] = 0$ .

Case A: $sin (θ /2) = 0 \Rightarrow cos θ = 1$ , so abscissa of $P = 2$ .

Case B: $tan (θ /2) = 5 \Rightarrow cos θ = \frac{1 - 25}{1 + 25} = - \frac{12}{13}$ , so abscissa $= 2 \cdot (- \frac{12}{13}) = - \frac{24}{13}$ .

Sum of abscissas $= 2 - \frac{24}{13} = \frac{26 - 24}{13} = \frac{2}{13}$ . Therefore $13 \times$ (sum) $= 2$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →