If circle (x−6)2+y2=r2 and parabola y2=4x have  maximum number of common chords, then least integral value of r is

For maximum number of common chords, circle and parabolamust 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 (x−6)2+4x=r2  or  x2−8x+36−r2=0 Curves touch if discriminant is 0 . D=64−436−r2=0 or r2=20Hence least integral value of r for which the curves intersect is 5.