Question

Since the expression (x-4)^2+(y+1)^2 evaluated at (3,6) is

Answer 1

$\bf (x-4)^2+(y+1)^2\qquad (3,6)~~ \begin{cases} x = 3\\ y = 6 \end{cases}\implies (3-4)^2+(6+1)^2 \\\\\\ (-1)^2+7^2\implies 1+49\implies 50$