Question

Suppose the circles \(x^{2}+y^{2}=1\) and \((x-1)^{2}+(y-1)^{2}=r^{2}\) intersect each other orthogonally at the point \((u,v)\). Then \(u+v=\) \(\underline{\hspace{1cm}}\).
 

Hint:

Orthogonal circles satisfy the condition that the dot product of the gradients of their equations at the intersection point is zero.

Answer 1

Correct Answer: 1

Two circles intersect orthogonally if their gradients at the point of intersection are perpendicular.
The first circle is centered at \((0,0)\) with radius \(1\).
The second circle is centered at \((1,1)\).
Orthogonality condition:
\[ 2x(u) + 2y(v) = 2[(u-1) + (v-1)] \] Simplifying:
\[ u + v = 1. \] Thus, the required value of \(u+v\) is \(1\).