Two conditions, and the curve has to contain the point AB & BC

Differentiating $x^{2} + y^{2} = 25$ with respect to $x$, remembering that $y$ is a function of $x$, gives $2x + 2y\frac{dy}{dx} = 0$, so

$$\frac{dy}{dx} = -\frac{x}{y}.$$

Two things about that expression are new and both matter. It needs both coordinates to produce a number, so a slope is attached to a point rather than to an $x$. And it is a quotient, so it has two ways of doing something interesting: the numerator can vanish, and the denominator can.

1. Horizontal tangent: numerator $= 0$ with denominator $\neq 0$.
2. Vertical tangent: denominator $= 0$ with numerator $\neq 0$.
3. Both zero: the test is inconclusive, and the point usually deserves separate attention. On the folium $x^{3} + y^{3} = 6xy$ that happens at the origin, where the curve crosses itself.

For the circle: the numerator $-x$ vanishes when $x = 0$ and the denominator $y$ vanishes when $y = 0$, so horizontal tangents live where $x = 0$ and vertical ones where $y = 0$. Which is exactly what a circle looks like, at the top, the bottom, and the two sides.

"Horizontal tangents where $x = 0$" is half an answer. The question asks where, meaning at which points, and a point has to be on the curve.

So substitute the condition back into the original equation. With $x = 0$ in $x^{2} + y^{2} = 25$: $y^{2} = 25$, so $y = \pm 5$, and the points are $(0, 5)$ and $(0, -5)$. The point $(0, 3)$ satisfies $x = 0$ and is not on the circle, since $0 + 9 \neq 25$, so it is not a point of tangency to anything.

This is the step that disappears most often, because the first condition feels like the whole calculation. It is worth doing on a curve where the second condition does real work. Take $x^{2} + xy + y^{2} = 3$, for which

$$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}.$$

1. Horizontal. $2x + y = 0$, so $y = -2x$. Substituting into the original: $x^{2} - 2x^{2} + 4x^{2} = 3x^{2} = 3$, so $x = \pm 1$ and the points are $(1, -2)$ and $(-1, 2)$.
2. Vertical. $x + 2y = 0$, so $x = -2y$. Substituting: $4y^{2} - 2y^{2} + y^{2} = 3y^{2} = 3$, so $y = \pm 1$ and the points are $(-2, 1)$ and $(2, -1)$.

The condition alone described a whole line in each case, and the original equation cut it down to two points. Both steps are required, in that order.

A circle is not the graph of a function, and neither is $x^{2} + xy + y^{2} = 3$. Three consequences.

1. A slope needs a point, not an input. "The slope where $x = 3$" is ambiguous on the circle: at $(3, 4)$ it is $-\frac{3}{4}$ and at $(3, -4)$ it is $+\frac{3}{4}$.
2. The same condition can produce several points. Both $(0, 5)$ and $(0, -5)$ have horizontal tangents, and they are the top and the bottom of the circle. Reporting one is reporting half the answer.
3. A relation can have no tangent at all somewhere. Where the curve crosses itself, as the folium does at the origin, there are two branches through one point and no single tangent line.

The vertical tangent points are worth a second look for a different reason: they are exactly the places where the relation fails to be a function locally, since just to the left or right of $(5, 0)$ on the circle there are two nearby $y$ values or none. That is why the derivative blows up there rather than merely being large.

Concavity works the same way it always has: it is the sign of $\frac{d^{2}y}{dx^{2}}$. Getting that quantity takes one extra move.

Differentiate $\frac{dy}{dx} = -\frac{x}{y}$ with the quotient rule, treating $y$ as a function of $x$:

$$\frac{d^{2}y}{dx^{2}} = -\frac{(1)(y) - (x)\frac{dy}{dx}}{y^{2}}.$$

Now substitute $\frac{dy}{dx} = -\frac{x}{y}$ back in, which is the step that gets skipped:

$$= -\frac{y + \frac{x^{2}}{y}}{y^{2}} = -\frac{y^{2} + x^{2}}{y^{3}} = -\frac{25}{y^{3}},$$

using the original equation at the last step. An answer still containing $\frac{dy}{dx}$ is unfinished, and one that treated $y$ as a constant, producing $-\frac{1}{y}$, is wrong.

The result is worth reading. $-\frac{25}{y^{3}}$ is negative when $y > 0$ and positive when $y < 0$, so the upper semicircle is concave down and the lower is concave up. Both halves curve toward the centre, and as graphs they have opposite concavity, which is what the second derivative reports. And as always, the sign of $\frac{dy}{dx}$ has nothing to do with it: on the arc from $(0, 5)$ to $(5, 0)$ the slope is negative and the concavity is down, while on the arc from $(-5, 0)$ to $(0, -5)$ the slope is negative and the concavity is up.