To find the locus of the center of the variable circle that intersects the given circles orthogonally, we need to find the equation of the locus.

Let's start by finding the equations of the given circles:

Circle 1: x^2 + y^2 = 4
Circle 2: x^2 + y^2 - 2x - 4y = 6

We can rewrite the second equation as:
x^2 - 2x + y^2 - 4y = 6

Now, let's consider a variable circle with center (h, k) and radius r. The equation of this circle can be written as:

(x-h)^2 + (y-k)^2 = r^2

To find the locus of the center of this variable circle, we need to find the values of h, k, and r that satisfy the condition of orthogonality with the given circles.

For two circles to be orthogonal, the product of their slopes must be -1. The slope of a circle at any point (x, y) can be found by differentiating the equation of the circle with respect to x and solving for dy/dx.

Let's find the slopes of the two given circles:

Circle 1: x^2 + y^2 = 4
Differentiating both sides with respect to x:
2x + 2y(dy/dx) = 0
dy/dx = -x/y

Circle 2: x^2 + y^2 - 2x - 4y = 6
Differentiating both sides with respect to x:
2x + 2y(dy/dx) - 2 - 4(dy/dx) = 0
dy/dx = (2x - 2)/(2 + 4y)

Now, let's find the values of h, k, and r that satisfy the orthogonality condition with the given circles.

For Circle 1:
(-h/-k) * (-x/y) = -1
h/k = x/y

For Circle 2:
(-h/-k) * ((2x - 2)/(2 + 4y)) = -1
h/k = (2x - 2)/(2 + 4y)

By equating the two expressions for h/k, we can eliminate the variable x:

x/y = (2x - 2)/(2 + 4y)

Simplifying this equation, we get:
2xy + 4y^2 = 2x^2 - 2x

Now, let's solve this equation to find the locus.

2xy + 4y^2 = 2x^2 - 2x
2x^2 - 2xy - 2x + 4y^2 = 0
x^2 - xy - x + 2y^2 = 0

Comparing this equation with the given equation of the locus, C1x + C2y = 1, we can see that:
C1 = -1
C2 = 2

Therefore, the value of (C1 C2) is (2).