Suppose the circles x² + y² = 1 and (x - 1)² + (y - 1)² = r² intersect each other orthogonally at the point (u, v). Then u + v = ______.

The correct answer is 1.

Let's analyze the given equations of the circles:
First circle (Curve 1):
$x^2+y^2=1$

Second circle (Curve 2):
${(x-1)}^2+{(y-1)}^2=r^2$

Both circles intersect orthogonally at the point $(u,v)$. This means that the tangents to the two circles at the point of intersection $(u,v)$ are perpendicular to each other.

First, since $(u,v)$ lies on the first circle, its coordinates must satisfy the equation of the first circle:
$u^2+v^2=1$

Now, let's find the slopes of the tangents to the circles at the point $(u,v)$ by differentiation.

Differentiating Curve 1 with respect to $x$:
$2x+2y\frac{dy}{dx}=0$
This gives the slope of the tangent $m_1$ at the point $(u,v)$:
$m_1=-\frac{u}{v}$

Differentiating Curve 2 with respect to $x$:
$2(x-1)+\mathrm{2<!--="true"-->(\; y}-1)\frac{dy}{dx}=0$
This gives the slope of the tangent $m_2$ at the point $(u,v)$:
$m_2=-\frac{u-1}{v-1}=\frac{1-u}{v-1}$

Since the circles intersect orthogonally at $(u,v)$, the product of the slopes of the tangents is $-1$:
$m_1\cdot m_2=-1$

Substituting the values of $m_1$ and $m_2$:
$\left(-\frac{u}{v}\right)\cdot \left(\frac{1-u}{v-1}\right)=-1$

Multiplying the terms:
$\frac{-u(1-u)}{v(v-1)}=-1$
Simplifying the negative signs:
$\frac{u-u^2}{v^2-v}=-1$

Cross-multiplying gives:
$u-u^2=-(v^2-v)$
$u-u^2=-v^2+v$
Rearranging the equation:
$u^2+v^2=u+v$

Since we already established that $u^2+v^2=1$, we substitute this value into the equation:
$1=u+v$

Thus, we obtain:
$u+v=1$