JEE Main 2026 — Circle Question with Solution

Circle $C_1$ has radius 5 and chord on line $y=x$ through origin.
Setting center at $(5,0)$ (in region $x\ge 0$), the chord endpoints are $(0,0)$ and $(5,5)$ by solving $(t-5{)}^2+t^2=25$.
Circle $C_2$ has this chord as diameter, so center is at $(\frac{5}{2},\frac{5}{2})$ with radius $\frac{5\sqrt{2}}{2}$.
The chord of $C_2$ through $(2,3)$ that is farthest from center is perpendicular to the radial direction from center to $(2,3)$.
The radial direction is $(2-\frac{5}{2},3-\frac{5}{2})=(-\frac{1}{2},\frac{1}{2})$ with normal direction $(1,-1)$.
The chord equation is $1(x-2)-1(y-3)=0$, giving $x-y+1=0$.
In form $x+ay+b=0$, we have $a=-1$ and $b=1$, so $a-b=-1-1=-2$.