JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Circle

The circle $C_1$ has radius $5$, passes through the origin and lies in $x\ge 0$, so its centre is $(5,0)$ and $C_1:(x-5{)}^2+y^2=25$. The chord $y=x$ meets $C_1$ where $(x-5{)}^2+x^2=25$, i.e. $2x^2-10x=0$, giving the endpoints $(0,0)$ and $(5,5)$. Taking this chord as a diameter, $C_2$ has centre $\left(\frac{5}{2},\frac{5}{2}\right)$ and equation $x^2+y^2-5x-5y=0$. Among the chords of $C_2$ through $(2,3)$, the one farthest from the centre is perpendicular to the join of the centre to $(2,3)$; that join has slope $\frac{3-5/2}{2-5/2}=-1$, so the required chord has slope $1$. Its equation is $y-3=x-2$, i.e. $x-y+1=0$. Comparing with $x+ay+b=0$ gives $a=-1$ and $b=1$, so $a-b=-2$.