Which point lies on a circle with a radius of 5 units and center at p(6, 1)?
a) q(1, 11)
b) r(2, 4)
c) s(4, -4)
d) t(9, -2)

An equation of a circle:

(x - a)² + (y - b)² = r²

(a; b) - a coordinates of a center
r - a radius

r = 5; P(6; 1) ⇒ a = 6 and b = 1

therefore

(x - 6)² + (y - 1)² = 5²

check:
A) Q(1; 11)
L = (1 - 6)² + (11 - 1)² = (-5)² + 10² = 25 + 100 = 125
R = 25
L ≠ R

B) R(2; 4)
L = (2 - 6)² + (4 - 1)² = (-4)² + 3² = 16 + 9 = 25
R = 25
L = R - CORRECT

C) S(4; -4)
L = (4 - 6)² + (-4 - 1)² = (-2)² + (-5)² = 4 + 25 = 29
R = 25
L ≠ R

D) T(9; -2)
L = (9 - 6)² + (-2 - 1)² = 3² + (-3)² = 9 + 9 = 18
R = 25
L ≠ R

B) R(2; 4).

The general equation for a circle centered at (h,k) with radius, R, is 
(x-h)^2 + (y-k)^2 = R^2

Putting the values in formula,
(x-6)^2 + (y-1)^2 = 5^2
Putting the coordinates (2, 4) of the answer B(2-6)^2 + (4-1)^2 = 25
(-4)^2 + 3^2 = 25
16 + 9 = 25
25 = 25
R.H.S = L.H.S
So according to above solution,

B) R(2, 4), is the correct answer.