JEE Main · 2016 · 03 Apr 2016 · Offline · Parabola

Write a point on $y^2=8x$, where $a=2$, as $(2t^2,4t)$. The shortest distance to the circle's centre $C(0,-6)$ is along the parabola's normal through $C$. That normal is $y=-tx+4t+2t^3$; putting in $C(0,-6)$ gives $-6=4t+2t^3$, i.e. $t^3+2t+3=0$, whose real root is $t=-1$, so $P=(2,-4)$. The required circle is centred at $P$ and passes through $C$, so its radius is $PC=\sqrt{2^2+(-4+6{)}^2}=2\sqrt{2}$. Thus $(x-2{)}^2+(y+4{)}^2=8$, which expands to $x^2+y^2-4x+8y+12=0$.