Parabolas and circles: equations, centres and vertices

The circle: every point at one distance from the centre

A circle with centre $C(x_0, y_0)$ and radius $r$ is the set of points $(x, y)$ at distance $r$ from $C$. Squaring the distance formula gives its equation:

$$(x - x_0)^2 + (y - y_0)^2 = r^2$$

Watch the signs: $(x + 3)^2 + (y - 1)^2 = 4$ has centre $(-3, 1)$ and radius $2$, because $x + 3 = x - (-3)$. And the right-hand side is $r^2$, so the radius is its square root.

The expanded form

Multiplying out gives the form questions usually use:

$$x^2 + y^2 + ax + by + c = 0$$

Read the centre and radius from it with two rules:

$$\text{centre} = \left(-\frac{a}{2}, -\frac{b}{2}\right) \qquad r^2 = \frac{a^2}{4} + \frac{b^2}{4} - c$$

Both come from completing the square. For $x^2 - 6x$, half of $-6$ is $-3$, and $x^2 - 6x = (x - 3)^2 - 9$. Doing the same in $y$ and moving the constants across gives the centre form. Completing the square is slower than the formulas but works when you have forgotten them, so practise both.

Three conditions make an equation of this kind a circle:

• the coefficients of $x^2$ and $y^2$ are equal (if both are 2, divide the whole equation by 2 first);
• there is no $xy$ term;
• the computed $r^2$ is positive. If $r^2 = 0$ the "circle" is a single point; if $r^2 < 0$ no point satisfies the equation at all.

A circle with centre $(x_0, y_0)$ touches the $x$-axis exactly when $\lvert y_0 \rvert = r$, and touches the $y$-axis when $\lvert x_0 \rvert = r$. It passes through the origin exactly when $c = 0$.

The parabola with a vertical axis

The graph of $y = ax^2 + bx + c$ with $a \neq 0$ is a parabola whose axis of symmetry is vertical. Picture a U shape: it opens upwards if $a > 0$ (the vertex is the lowest point) and downwards if $a < 0$ (the vertex is the highest point). The larger $\lvert a \rvert$, the narrower the U.

The vertex has

$$x_V = -\frac{b}{2a}$$

and its $y$-coordinate comes from substituting $x_V$ into the equation. The axis of symmetry is the vertical line $x = x_V$. Other landmarks follow at once:

• the $y$-intercept is $(0, c)$;
• the $x$-intercepts solve $ax^2 + bx + c = 0$; there are two if the discriminant $\Delta = b^2 - 4ac$ is positive, one (the vertex itself touches the axis) if $\Delta = 0$, and none if $\Delta < 0$;
• the two $x$-intercepts, when they exist, sit symmetrically about the axis, so their average is $x_V$.

Vertex form

Completing the square writes the parabola as $y = a(x - h)^2 + k$, with vertex $(h, k)$. This form is the fastest way to build a parabola from its vertex and one more point: substitute the point and solve for $a$.

A parabola can also have a horizontal axis, $x = ay^2 + by + c$. Every rule above holds with $x$ and $y$ swapped: it opens to the right if $a > 0$, its vertex has $y_V = -\frac{b}{2a}$, and it is not the graph of a function of $x$.

Building equations from conditions

Count the unknowns. A circle in expanded form has three ($a$, $b$, $c$), so three points on it fix it; so do a centre and a radius, or the two ends of a diameter (the midpoint is the centre, half the length is the radius). A parabola $y = ax^2 + bx + c$ also has three, so three points fix it, or a vertex (two conditions) and one more point. Always check your final equation with a given point; it costs seconds and catches sign slips.