Introduction:
We are given two equations: x^2 + y^2 = 4 (Equation 1) and x^2 + y^2 - 6x - 8y + 24 = 0 (Equation 2). We need to find the number of circles that can be drawn to touch at least three of the common tangents.

Step 1: Find the Center of the Circle:
To find the center of the circle, we need to convert Equation 2 into the standard form of a circle equation, (x - h)^2 + (y - k)^2 = r^2.

Rearranging Equation 2, we get:
x^2 + y^2 - 6x - 8y + 24 = 0
x^2 - 6x + y^2 - 8y = -24

Completing the square for x and y terms, we have:
(x^2 - 6x + 9) + (y^2 - 8y + 16) = -24 + 9 + 16
(x - 3)^2 + (y - 4)^2 = 1

Comparing this with the standard form of a circle equation, we can conclude that the center of the circle is (3, 4) and the radius is 1.

Step 2: Find the Common Tangents:
The common tangents to two circles are the lines that are tangent to both circles at the same point. To find the common tangents, we need to solve the system of equations formed by Equations 1 and 2.

Substituting Equation 1 into Equation 2, we get:
4 - 6x - 8y + 24 = 0
-6x - 8y + 28 = 0
3x + 4y - 14 = 0

This equation represents a straight line, which is one of the common tangents.

To find the second common tangent, we differentiate both equations with respect to x and eliminate the derivative to obtain another equation. Solving this equation with Equation 1 will give us the second common tangent.

Differentiating Equation 1 with respect to x, we get:
2x + 2yy' = 0
y' = -x/y

Similarly, differentiating Equation 2 with respect to x, we get:
2x - 6 + 2yy' - 8y' = 0
2x + 2yy' - 8y' = 6
2x + 2y(-x/y) - 8(-x/y) = 6
2x - 2x + 8 = 6
8 = 6

This equation is inconsistent, which means there is no second common tangent. Therefore, there is only one common tangent.

Step 3: Find the Circles:
To find the circles that can be drawn to touch at least three of the common tangents, we need to find the circles that are tangent to the given common tangent.

The tangent to a circle is perpendicular to the radius at the point of contact. Therefore, the circles that are tangent to the common tangent will have their centers lying on the perpendicular bisector of the common tangent.

The perpendicular bisector of the