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

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

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(x-6{)}^2+4x=r^2\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}^2-8x+36-r^2=0\end{array}$

Curves touch if discriminant is 0.

$D=64-4\left(36-r^2\right)=0\text{ or }{r}^2=20$

Hence least integral value of r for which the curves intersect is 5.