Answer
$F=-y i +x j$
Work Step by Step
We will parameterize the equation of the circle as follows:
$r(t) =\sqrt {a^2+b^2} \cos t i +\sqrt {a^2 +b^2} \sin t j ...(1)$
Now, differentiate the equation (1) to find the tangent vector $T(t)$.
So, $T(t)=- \sqrt {a^2+b^2} \sin t i +\sqrt {a^2 +b^2} \cos t j$
or, $T =-y i +x j$
Thus, the equation for the vector field becomes: $F=-y i +x j$
