JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Circle

That $x-y+1=0$ is the perpendicular bisector of $PQ$ means $Q$ is the mirror image of $P$ in this line. Reflecting $P=(x_0,y_0)$ in $x-y+1=0$ gives $Q=(y_0-1,x_0+1)$. Since $Q$ lies on $5x+y+2=0$, $5(y_0-1)+(x_0+1)+2=0$, i.e. $x_0+5y_0-2=0$, so $x_0=2-5y_0$. As $P$ lies on $x^2+y^2=4$, $(2-5y_0{)}^2+y_0^2=4$, giving $26y_0^2-20y_0=0$, so $y_0=0$ or $y_0=\frac{10}{13}$. The corresponding abscissae are $x_0=2$ and $x_0=-\frac{24}{13}$, whose sum is $\frac{2}{13}$. Thus $13$ times the sum of the abscissae equals $2$.