Question

The coordinates of the point on the parabola $x^2=8y$, whose distance from the circle $(x-4{)}^2+(y-2{)}^2=1$ is minimum is

• A. $0$
• B. $2$
• C. $3$
• D. $5$

Answer 1

Answer: (A)

$0$

The centre of the circle is $(4,2)$ which lies on $x^2=8y$
$\therefore$ Required distance is zero .