JEE Main 2026 — Jan 28, Shift 1 — Mathematics solution

JEE Main2026Jan 28, Shift 1Mathematics

Let $y = x$ be the equation of a chord of the circle $C_{1}$ (in the closed half-plane $x \geq 0$ ) of diameter 10 passing through the origin. Let $C_{2}$ be another circle described on the given chord as its diameter. If the equation of the chord of the circle $C_{2}$ , which passes through the point $(2, 3)$ and is farthest from the center of $C_{2}$ , is $x + a y + b = 0$ , then $a - b$ is equal to:

A
10
B
$- 6$
C
$- 2$
D
$6$

Solution

$C_{1}$ has diameter 10, so radius 5, and the chord $y = x$ passes through the origin. The chord of $C_{1}$ along $y = x$ with length 10 (a diameter chord through O) has endpoints $(0, 0)$ and $(5, 5)$ in the half-plane $x \geq 0$ .

$C_{2}$ is described on this chord as diameter, so its endpoints are $(0, 0)$ and $(5, 5)$ . Centre of $C_{2}$ is the midpoint $A = (\frac{5}{2}, \frac{5}{2})$ .

The chord of $C_{2}$ that is farthest from its centre and passes through $B = (2, 3)$ is the one perpendicular to $A B$ at $B$ .

Slope of $A B$ : $m_{A B} = \frac{3 - 5/2}{2 - 5/2} = \frac{1/2}{- 1/2} = - 1$ , so the required chord has slope $+ 1$ .

Equation through $(2, 3)$ : $y - 3 = 1 \cdot (x - 2) \Rightarrow x - y + 1 = 0$ .

Comparing with $x + a y + b = 0$ : $a = - 1$ , $b = 1$ , hence $a - b = - 2$ .

Option (3).

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