How do I do this question on circles? 10 points

Two circles, which do not touch or overlap, have as their equations:

(x - 15)^2 + (y - 6)^2 = 40 and x^2 + y^2 - 6x -4y + 3 = 0

Show that the shortest distance between the two circles is equal to the radius of the smaller circle if the exact distance between the centres of the two circles is 4√10.

It's a non-cal paper. Thanks in advance

1st find eq. of 2nd circle by completing the square for "x" and for "y".

x² - 6x + ___ + y² - 4y + ___ = -3 Recall: A² + 2AB + B² = (A + B)² so the "x's":-6/2= -3 and (-3)²=9,
x² - 6x + 9 + y² - 4y + 2 = -3 + 9 + 4....and for the "y's": -4/2= -2 and (-2)² = 4
(x - 6)² + (y - 2)² = 10
This circle has a center at (3, 2) and a radius of √10

(x - 15)² + (y - 6)² = 40
This circle has a center at (15, 6) and a radius of √40 = √4*√10 = 2√10

If the distance between the centers is 4√10 and since they can't touch we subtract the radius of each of our circles we get:
4√10 - 2√10 - 1√10 = √10 which is the radius of our smaller circle.

centre c1 =(15,6) r1= 2√10. c2 =(3,2) r2 = √10.
distance between c1and c2 = 4√10 =2√10 + shortest distance between the two circles+ shortest distance between the two circles
shortest distance between the two circles = √10. = r2