JEE Main 2026 — Mathematics Coordinate Geometry

Circle $C_1$C1​ has radius 5 and chord on line $y = x$y=x through origin.

Setting center at $(5, 0)$(5,0) (in region $x \geq 0$x≥0), the chord endpoints are $(0, 0)$(0,0) and $(5, 5)$(5,5) by solving $(t-5)^2 + t^2 = 25$(t−5)2+t2=25.

Circle $C_2$C2​ has this chord as diameter, so center is at $(\frac{5}{2}, \frac{5}{2})$(25​,25​) with radius $\frac{5\sqrt{2}}{2}$252​​.

The chord of $C_2$C2​ through $(2, 3)$(2,3) that is farthest from center is perpendicular to the radial direction from center to $(2, 3)$(2,3).

The radial direction is $(2 - \frac{5}{2}, 3 - \frac{5}{2}) = (-\frac{1}{2}, \frac{1}{2})$(2−25​,3−25​)=(−21​,21​) with normal direction $(1, -1)$(1,−1).

The chord equation is $1(x - 2) - 1(y - 3) = 0$1(x−2)−1(y−3)=0, giving $x - y + 1 = 0$x−y+1=0.

In form $x + ay + b = 0$x+ay+b=0, we have $a = -1$a=−1 and $b = 1$b=1, so $a - b = -1 - 1 = -2$a−b=−1−1=−2.