Question

If circle $(x-6{)}^2+y^2=r^2$ and parabola $y^2=4x$ have maximum number of common chords, then least integral value of $r$ is _____

Answer 1

Answer: 5

For maximum number of common chords, circle and parabola must intersect in $4$ distinct points.
Let's first find the value of $r$ when circle and parabola touch each other.
For that solving the given curves we have
or $(x-6{)}^2+4x=r^2$
$x^2-8x+36-r^2=0$
Curves touch if discriminant is 0 .
$D=64-4\left(36-r^2\right)=0$ or $r^2=20$
Hence least integral value of $r$ for which the curves intersect is $5.$