It does not follow; for a counterexample use the construction \[ y = x^2 \qquad\text{and}\qquad (x+4)^2 + \left( y - \frac 72 \right)^2 = \frac{125}{4}. \]See figure below. They meet at $A = (1,1)$ and $B = (-3,9)$. [asy][asy] import graph; real f(real x) { return x*x; } draw(graph(f,-3.5,3,operator ..), blue, Arrows); draw(circle((-4, 7/2), (125/4.0)**.5), red); dot("$A$", (1,1), dir(315)); dot("$B$", (-3,9), dir(45)); [/asy][/asy] Remark: [Motivation for these numbers] If we write the parabola as $y=x^2$ WLOG, and the circle as $(x-\alpha)^2 + (y-\beta)^2 = \gamma^2$, then the interesctions are controlled by a quartic in $x$, namely: \begin{align*} 0 &= \left( x-\alpha \right)^2 + \left( x^2 - \beta \right)^2 - \gamma^2 \\ &= x^4 + (1-2\beta) x^2 - 2\alpha x + (\alpha^2+\beta^2-\gamma^2). \end{align*}To fail the problem, we need this quartic to have a ``triple root'' at $A$. The above example is constructed by choosing $(x-1)^3(x+3)$ as the quartic and equating coefficients.

The answer is no. Suppose the parabola is $y=x^2$. Now, suppose the circle is $(x-a)^2+(y-b)^2=r^2$. If we plug in $y=x^2$, we get a quartic in $x$ that is $$x^4+x^2(1-2b)+x(-2a)+(a^2+b^2-r^2)=0 (*)$$ The idea is that it is possible for the quartic to have a triple root at $A$ and a single root at $B$. Suppose the roots are $s,s,s,-3s$. Consider $s=1$, so $$(x-1)^3(x-3)=x^4-6x^2+8x-3.$$Equating the coefficents in $(*)$, we can get that $b=\frac{7}{2}$, $a=-4$, and $r=\frac{5\sqrt{5}}{2}$. This gives an example for which the problem statement is false, specifically when the parabola is $y=x^2$ and the circle is centered at $(-4,\frac{7}{2})$ with radius $\frac{5\sqrt{5}}{2}$, as now there is a triple root at $A$ and only a single root at $B$.