= Solution
{c}
A <streamline> is a curve whose tangent is parallel to the instantaneous fluid <velocity field>. In steady flow one may parametrize it by $d\mathbf r/ds=\mathbf u(\mathbf r)$. Dotting the steady <Euler equations> with $\mathbf u$ eliminates the cross-product term, since $\mathbf u\cdot((\nabla\times\mathbf u)\times\mathbf u)=0$. Hence
$$
\boxed{\mathbf u\cdot\nabla\left(p+\tfrac12\rho|\mathbf u|^2\right)=0,}
$$
which is <Bernoulli's principle> along each streamline.
For the supplied field, $xu+yv=0$, so along a streamline
$$
\frac{d}{ds}(x^2+y^2)=2xu+2yv=0.
$$
Writing $r^2=x^2+y^2$, the tangential speed is $v_\theta=-br+c/r$ and the radial speed is zero. Both $r$ and $|\mathbf u|^2=(-br+c/r)^2$ are constant around each nonstagnant circular streamline. The Bernoulli quantity is constant there, so \b[the pressure is constant on every such circle].
If a circle is wholly stagnant, the same conclusion follows directly from the steady <Euler equations>: on it $\mathbf u=0$ and $\nabla|\mathbf u|^2=0$, hence $\nabla p=0$. Equivalently the radial momentum equation gives $\nabla p=\rho(-br+c/r)^2\mathbf e_r/r$ and no angular pressure variation, including at zero tangential speed. Thus \b[$p$ is constant on $x^2+y^2=a^2$ for every nonzero $a$], with no exception for a stagnation circle.
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