Lines and conics: secant, tangent or apart

Three possible positions

A straight line and a circle can meet in two points (the line is a secant), in exactly one point (the line is tangent, touching the circle without crossing it), or not at all (the line is external). The same three cases describe a line and a parabola, with one exception covered below. Questions ask either for the position, for the common points themselves, or for the value of a parameter that produces a given position.

Method 1: substitute and look at the discriminant

The common points satisfy both equations, so solve the system. Solve the line for one variable (usually $y$), substitute into the conic, and collect everything on one side. You get a second-degree equation in one unknown, the resolvent equation. Its discriminant $\Delta$ decides the position:

Discriminant	Common points	Position
$\Delta > 0$	two	secant
$\Delta = 0$	one (a double root)	tangent
$\Delta < 0$	none	external

If the question asks for the points, solve the resolvent equation and put each root back into the line's equation to get the other coordinate. The line is simpler, and each root gives exactly one point.

Only the discriminant's sign matters for the position, so you can often stop before solving. When the $x$-coefficient $b$ is even, the reduced discriminant $\frac{\Delta}{4} = \left(\frac{b}{2}\right)^2 - ac$ has the same sign and smaller numbers.

Method 2 for circles: distance against radius

For a circle there is a quicker route. Compute the distance $d$ from the centre to the line with the point-to-line formula from the lesson on points, distances and lines, then compare it with the radius:

• $d < r$: secant;
• $d = r$: tangent;
• $d > r$: external.

This avoids expanding squares, and it is the natural method for questions such as "for which radius is this circle tangent to this line?". When $d$ and $r$ are both roots, compare $d^2$ and $r^2$ instead. Method 2 tells you the position but not the points; use method 1 when the points are asked for.

The tangent point itself is the foot of the perpendicular from the centre to the line. A tangent is always perpendicular to the radius at the point of contact, so the tangent at a point $T$ of a circle with centre $C$ is the line through $T$ perpendicular to $CT$.

Parabolas: the vertical line exception

For a parabola $y = ax^2 + bx + c$ there is no centre, so method 1 is the tool. One case needs care. A vertical line $x = h$ meets this parabola in exactly one point, $(h, ah^2 + bh + c)$, but it is not tangent: it crosses the curve, parallel to the axis. Substitution here gives no quadratic at all, so the discriminant test does not apply. Only non-vertical lines are tangent when $\Delta = 0$.

Parameters: tangent lines and families

Many questions contain a letter: "for which $q$ is the line $y = 2x + q$ tangent to the parabola?" Substitute, form the resolvent equation with $q$ still in it, and set $\Delta = 0$ (tangent), $\Delta > 0$ (two points) or $\Delta < 0$ (none). The result is an equation or inequality in the parameter.

Think about the picture to check the direction of an inequality. Lines $y = 2x + q$ are all parallel; as $q$ grows they slide upwards. For an upward-opening parabola, high lines cut it in two points and low lines miss it, so "two points" must come out as $q$ greater than the tangent value. If your inequality points the other way, look for a sign slip.

A routine that works every time

1. Write the line as $y = \ldots$ (or $x = \ldots$ if that is simpler).
2. Substitute into the conic and move every term to one side.
3. Simplify, and divide by any common factor.
4. Use the sign of $\Delta$ for the position, or the roots for the points.
5. Check one common point in both original equations.