The least distance of the chord passing through (2 , 1) of the circle x^2+y^2-2x-4y-13=0

A) 2 B) 6 C) 8 D) 4

x² + y² - 2x - 4y - 13 = 0
By completing square method , we have
(x -1)² + (y -2)² = 13 + 4 + 1

(x -1)² + (y -2)² = 18
So the radius is root(18) and centre is ( 1 , 2)

The point on the chord is  (2,1)
So the distance from centre to the chord is $\sqrt{(2-1{)}^2+(1-2{)}^2}=\sqrt{2}$
As the chord is passing through (2,1) , and root(2) is the distance which is less than then the answer options given .

{but the least distance is always the perpendicular distance }
So the answer options are wrong , or either your question has some data missing
Please check your question.